Solve 0=-x^2+x+12

$$0 = - x ^ { 2 } + x + 12$$

Solve for x

$$x=-3$$

$$x=4$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$-x^{2}+x+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=1 ab=-12=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=4 b=-3$$

Rewrite -x^{2}+x+12 as \left(-x^{2}+4x\right)+\left(-3x+12\right).

$$\left(-x^{2}+4x\right)+\left(-3x+12\right)$$

Factor out -x in the first and -3 in the second group.

$$-x\left(x-4\right)-3\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(-x-3\right)$$

To find equation solutions, solve x-4=0 and -x-3=0.

$$x=4 x=-3$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$-x^{2}+x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 1 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-1±\sqrt{1+4\times 12}}{2\left(-1\right)}$$

Multiply 4 times 12.

$$x=\frac{-1±\sqrt{1+48}}{2\left(-1\right)}$$

Add 1 to 48.

$$x=\frac{-1±\sqrt{49}}{2\left(-1\right)}$$

Take the square root of 49.

$$x=\frac{-1±7}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-1±7}{-2}$$

Now solve the equation x=\frac{-1±7}{-2} when ± is plus. Add -1 to 7.

$$x=\frac{6}{-2}$$

Divide 6 by -2.

$$x=-3$$

Now solve the equation x=\frac{-1±7}{-2} when ± is minus. Subtract 7 from -1.

$$x=-\frac{8}{-2}$$

Divide -8 by -2.

$$x=4$$

The equation is now solved.

$$x=-3 x=4$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$-x^{2}+x+12=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$-x^{2}+x=-12$$

Divide both sides by -1.

$$\frac{-x^{2}+x}{-1}=-\frac{12}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{1}{-1}x=-\frac{12}{-1}$$

Divide 1 by -1.

$$x^{2}-x=-\frac{12}{-1}$$

Divide -12 by -1.

$$x^{2}-x=12$$

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=12+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=12+\frac{1}{4}$$

Add 12 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{49}{4}$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{7}{2} x-\frac{1}{2}=-\frac{7}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=4 x=-3$$