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Quadratic Formula

The quadratic formula solves any quadratic equation, no exceptions. Write the equation in standard form ax2+bx+c=0ax^2 + bx + c = 0, and the solutions are x=−b±b2−4ac2ax = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Factoring only works when the numbers cooperate; the formula works every single time.

The price of that power is careful arithmetic. Almost every quadratic formula mistake is a sign slip — a −b-b handled wrong, or a negative cc dropped inside −4ac-4ac. This article walks through how to plug in cleanly, what the piece under the square root tells you before you finish, and four fully worked examples.

Identify a, b, and c first

Before touching the formula, get the equation into standard form: everything on one side, zero on the other, terms in order. In ax2+bx+c=0ax^2 + bx + c = 0, the letter aa is the coefficient of x2x^2, bb is the coefficient of xx, and cc is the constant — signs included.

For x2−5x+6=0x^2 - 5x + 6 = 0, that means a=1a = 1, b=−5b = -5, c=6c = 6. The most common error is writing b=5b = 5 and losing the negative. A good habit: write aa, bb, and cc down on paper, each with its sign, before substituting anything.

If the equation isn't in standard form — say 2x2=7x−32x^2 = 7x - 3 — move everything to one side first: 2x2−7x+3=02x^2 - 7x + 3 = 0. Only then can you read off a=2a = 2, b=−7b = -7, c=3c = 3.

The discriminant tells you what's coming

The expression under the square root, b2−4acb^2 - 4ac, is called the discriminant. Compute it first, on its own, and it previews the answer: if it's positive, there are two real solutions; if it's zero, exactly one; if it's negative, no real solutions, because you can't take the square root of a negative number in the real numbers.

The discriminant also tells you whether the answers will be tidy. If b2−4acb^2 - 4ac is a perfect square like 2525, the solutions are rational — and the equation would have factored. If it's not a perfect square, like 2828, expect square roots in your answer, and simplify the radical: 28=27\sqrt{28} = 2\sqrt{7}.

A clean substitution routine

Write the formula fresh every time: x=−b±b2−4ac2ax = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Then substitute with parentheses around every value: x=−(−5)±(−5)2−4(1)(6)2(1)x = \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)}. The parentheses look fussy, but they are what keep (−5)2=25(-5)^2 = 25 from turning into −25-25.

Simplify in this order: the discriminant under the root, then the square root itself, then the ±\pm split into two separate answers. Finish by checking at least one solution in the original equation — substitution takes seconds and catches nearly everything.

Worked examples

Example 1: two integer solutions

Solve x2+5x+6=0x^2 + 5x + 6 = 0.

Identify the coefficientsa=1,b=5,c=6a = 1, \quad b = 5, \quad c = 6
Compute the discriminantb2−4ac=25−24=1b^2 - 4ac = 25 - 24 = 1
Substitute into the formulax=−5±12x = \dfrac{-5 \pm \sqrt{1}}{2}
Split the ±\pm into two answersx=−5+12=−2,x=−5−12=−3x = \dfrac{-5 + 1}{2} = -2, \quad x = \dfrac{-5 - 1}{2} = -3
Check x=−2x = -2: (−2)2+5(−2)+6=4−10+6=0(-2)^2 + 5(-2) + 6 = 4 - 10 + 6 = 0 ✓

Answer: x=−2x = -2 or x=−3x = -3

Example 2: a leading coefficient bigger than 1

Solve 2x2−7x+3=02x^2 - 7x + 3 = 0.

Identify the coefficients — keep the sign on bba=2,b=−7,c=3a = 2, \quad b = -7, \quad c = 3
Compute the discriminant(−7)2−4(2)(3)=49−24=25(-7)^2 - 4(2)(3) = 49 - 24 = 25
Substitute — note −b=−(−7)=7-b = -(-7) = 7 and 2a=42a = 4x=7±254=7±54x = \dfrac{7 \pm \sqrt{25}}{4} = \dfrac{7 \pm 5}{4}
Split into two answersx=124=3,x=24=12x = \dfrac{12}{4} = 3, \quad x = \dfrac{2}{4} = \dfrac{1}{2}
Check x=3x = 3: 2(9)−7(3)+3=18−21+3=02(9) - 7(3) + 3 = 18 - 21 + 3 = 0 ✓

Answer: x=3x = 3 or x=12x = \dfrac{1}{2}

Example 3: irrational solutions

Solve x2−4x−3=0x^2 - 4x - 3 = 0.

Identify the coefficientsa=1,b=−4,c=−3a = 1, \quad b = -4, \quad c = -3
Compute the discriminant — watch the double negative(−4)2−4(1)(−3)=16+12=28(-4)^2 - 4(1)(-3) = 16 + 12 = 28
Not a perfect square, so simplify the radical28=4⋅7=27\sqrt{28} = \sqrt{4 \cdot 7} = 2\sqrt{7}
Substitutex=4±272x = \dfrac{4 \pm 2\sqrt{7}}{2}
Divide every term in the numerator by 22x=2±7x = 2 \pm \sqrt{7}

Answer: x=2+7x = 2 + \sqrt{7} or x=2−7x = 2 - \sqrt{7}

Example 4: a discriminant of zero

Solve x2−6x+9=0x^2 - 6x + 9 = 0.

Identify the coefficientsa=1,b=−6,c=9a = 1, \quad b = -6, \quad c = 9
Compute the discriminant(−6)2−4(1)(9)=36−36=0(-6)^2 - 4(1)(9) = 36 - 36 = 0
A zero discriminant means exactly one solutionx=6±02=62x = \dfrac{6 \pm \sqrt{0}}{2} = \dfrac{6}{2}
Simplifyx=3x = 3
Check: 32−6(3)+9=9−18+9=03^2 - 6(3) + 9 = 9 - 18 + 9 = 0 ✓

Answer: x=3x = 3

Try one yourself

Common questions

When should I use the quadratic formula instead of factoring?

Try factoring for about ten seconds. If two numbers that multiply to cc and add to bb don't jump out, switch to the formula — it always works, and grinding on a trinomial that doesn't factor is wasted time. Quick tell: if the discriminant b2−4acb^2 - 4ac isn't a perfect square, the equation doesn't factor over the integers.

What does it mean when the discriminant is negative?

The equation has no real solutions — the parabola never crosses the xx-axis. In Algebra 1, you answer 'no real solutions' and stop. In Algebra 2, a negative discriminant leads to complex solutions involving ii.

Why does the formula have a plus-or-minus sign?

A quadratic can cross the xx-axis at two points, and the ±\pm produces both of them: one solution uses +b2−4ac+\sqrt{b^2 - 4ac} and the other uses −b2−4ac-\sqrt{b^2 - 4ac}. When the discriminant is zero, adding and subtracting zero give the same value, which is why there's only one solution in that case.

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