Solve x^2+6x-52=3(x-8)

$$x ^ { 2 } + 6 x - 52 = 3 ( x - 8 )$$

Solve for x

$$x=-7$$

$$x=4$$

Steps Using Factoring

Use the distributive property to multiply 3 by x-8.

$$x^{2}+6x-52=3x-24$$

Subtract 3x from both sides.

$$x^{2}+6x-52-3x=-24$$

Combine 6x and -3x to get 3x.

$$x^{2}+3x-52=-24$$

Add 24 to both sides.

$$x^{2}+3x-52+24=0$$

Add -52 and 24 to get -28.

$$x^{2}+3x-28=0$$

To solve the equation, factor x^{2}+3x-28 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=3 ab=-28$$

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 -28.

$$-1,28 -2,14 -4,7$$

Calculate the sum for each pair.

$$-1+28=27 -2+14=12 -4+7=3$$

The solution is the pair that gives sum 3.

$$a=-4 b=7$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-4\right)\left(x+7\right)$$

To find equation solutions, solve x-4=0 and x+7=0.

$$x=4 x=-7$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 3 by x-8.

$$x^{2}+6x-52=3x-24$$

Subtract 3x from both sides.

$$x^{2}+6x-52-3x=-24$$

Combine 6x and -3x to get 3x.

$$x^{2}+3x-52=-24$$

Add 24 to both sides.

$$x^{2}+3x-52+24=0$$

Add -52 and 24 to get -28.

$$x^{2}+3x-28=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-28. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-28\right)=-28$$

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 -28.

$$-1,28 -2,14 -4,7$$

Calculate the sum for each pair.

$$-1+28=27 -2+14=12 -4+7=3$$

The solution is the pair that gives sum 3.

$$a=-4 b=7$$

Rewrite x^{2}+3x-28 as \left(x^{2}-4x\right)+\left(7x-28\right).

$$\left(x^{2}-4x\right)+\left(7x-28\right)$$

Factor out x in the first and 7 in the second group.

$$x\left(x-4\right)+7\left(x-4\right)$$

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

$$\left(x-4\right)\left(x+7\right)$$

To find equation solutions, solve x-4=0 and x+7=0.

$$x=4 x=-7$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3 by x-8.

$$x^{2}+6x-52=3x-24$$

Subtract 3x from both sides.

$$x^{2}+6x-52-3x=-24$$

Combine 6x and -3x to get 3x.

$$x^{2}+3x-52=-24$$

Add 24 to both sides.

$$x^{2}+3x-52+24=0$$

Add -52 and 24 to get -28.

$$x^{2}+3x-28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 3 for b, and -28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-28\right)}}{2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-28\right)}}{2}$$

Multiply -4 times -28.

$$x=\frac{-3±\sqrt{9+112}}{2}$$

Add 9 to 112.

$$x=\frac{-3±\sqrt{121}}{2}$$

Take the square root of 121.

$$x=\frac{-3±11}{2}$$

Now solve the equation x=\frac{-3±11}{2} when ± is plus. Add -3 to 11.

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

Divide 8 by 2.

$$x=4$$

Now solve the equation x=\frac{-3±11}{2} when ± is minus. Subtract 11 from -3.

$$x=-\frac{14}{2}$$

Divide -14 by 2.

$$x=-7$$

The equation is now solved.

$$x=4 x=-7$$

Steps for Completing the Square

Use the distributive property to multiply 3 by x-8.

$$x^{2}+6x-52=3x-24$$

Subtract 3x from both sides.

$$x^{2}+6x-52-3x=-24$$

Combine 6x and -3x to get 3x.

$$x^{2}+3x-52=-24$$

Add 52 to both sides.

$$x^{2}+3x=-24+52$$

Add -24 and 52 to get 28.

$$x^{2}+3x=28$$

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=28+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=28+\frac{9}{4}$$

Add 28 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{121}{4}$$

Factor x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{11}{2} x+\frac{3}{2}=-\frac{11}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=4 x=-7$$