Solve 0=-x^2+9x+52

$$0 = - x ^ { 2 } + 9 x + 52$$

Solve for x

$$x=-4$$

$$x=13$$

Steps Using Factoring By Grouping

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

$$-x^{2}+9x+52=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+52. To find a and b, set up a system to be solved.

$$a+b=9 ab=-52=-52$$

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 -52.

$$-1,52 -2,26 -4,13$$

Calculate the sum for each pair.

$$-1+52=51 -2+26=24 -4+13=9$$

The solution is the pair that gives sum 9.

$$a=13 b=-4$$

Rewrite -x^{2}+9x+52 as \left(-x^{2}+13x\right)+\left(-4x+52\right).

$$\left(-x^{2}+13x\right)+\left(-4x+52\right)$$

Factor out -x in the first and -4 in the second group.

$$-x\left(x-13\right)-4\left(x-13\right)$$

Factor out common term x-13 by using distributive property.

$$\left(x-13\right)\left(-x-4\right)$$

To find equation solutions, solve x-13=0 and -x-4=0.

$$x=13 x=-4$$

Steps Using the Quadratic Formula

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

$$-x^{2}+9x+52=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 9 for b, and 52 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\left(-1\right)\times 52}}{2\left(-1\right)}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\left(-1\right)\times 52}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-9±\sqrt{81+4\times 52}}{2\left(-1\right)}$$

Multiply 4 times 52.

$$x=\frac{-9±\sqrt{81+208}}{2\left(-1\right)}$$

Add 81 to 208.

$$x=\frac{-9±\sqrt{289}}{2\left(-1\right)}$$

Take the square root of 289.

$$x=\frac{-9±17}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-9±17}{-2}$$

Now solve the equation x=\frac{-9±17}{-2} when ± is plus. Add -9 to 17.

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

Divide 8 by -2.

$$x=-4$$

Now solve the equation x=\frac{-9±17}{-2} when ± is minus. Subtract 17 from -9.

$$x=-\frac{26}{-2}$$

Divide -26 by -2.

$$x=13$$

The equation is now solved.

$$x=-4 x=13$$

Steps for Completing the Square

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

$$-x^{2}+9x+52=0$$

Subtract 52 from both sides. Anything subtracted from zero gives its negation.

$$-x^{2}+9x=-52$$

Divide both sides by -1.

$$\frac{-x^{2}+9x}{-1}=-\frac{52}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{9}{-1}x=-\frac{52}{-1}$$

Divide 9 by -1.

$$x^{2}-9x=-\frac{52}{-1}$$

Divide -52 by -1.

$$x^{2}-9x=52$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=52+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=52+\frac{81}{4}$$

Add 52 to \frac{81}{4}.

$$x^{2}-9x+\frac{81}{4}=\frac{289}{4}$$

Factor x^{2}-9x+\frac{81}{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}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{289}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{289}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{17}{2} x-\frac{9}{2}=-\frac{17}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=13 x=-4$$