Solve x^2+6x+9=144

$$x ^ { 2 } + 6 x + 9 = 144$$

Solve for x

$$x=-15$$

$$x=9$$

Steps Using Factoring

Subtract 144 from both sides.

$$x^{2}+6x+9-144=0$$

Subtract 144 from 9 to get -135.

$$x^{2}+6x-135=0$$

To solve the equation, factor x^{2}+6x-135 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.

$$a+b=6 ab=-135$$

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 -135.

$$-1,135 -3,45 -5,27 -9,15$$

Calculate the sum for each pair.

$$-1+135=134 -3+45=42 -5+27=22 -9+15=6$$

The solution is the pair that gives sum 6.

$$a=-9 b=15$$

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

$$\left(x-9\right)\left(x+15\right)$$

To find equation solutions, solve x-9=0 and x+15=0.

$$x=9 x=-15$$

Steps Using Factoring By Grouping

Subtract 144 from both sides.

$$x^{2}+6x+9-144=0$$

Subtract 144 from 9 to get -135.

$$x^{2}+6x-135=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-135. To find a and b, set up a system to be solved.

$$a+b=6 ab=1\left(-135\right)=-135$$

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 -135.

$$-1,135 -3,45 -5,27 -9,15$$

Calculate the sum for each pair.

$$-1+135=134 -3+45=42 -5+27=22 -9+15=6$$

The solution is the pair that gives sum 6.

$$a=-9 b=15$$

Rewrite x^{2}+6x-135 as \left(x^{2}-9x\right)+\left(15x-135\right).

$$\left(x^{2}-9x\right)+\left(15x-135\right)$$

Factor out x in the first and 15 in the second group.

$$x\left(x-9\right)+15\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(x+15\right)$$

To find equation solutions, solve x-9=0 and x+15=0.

$$x=9 x=-15$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x^{2}+6x+9=144$$

Subtract 144 from both sides of the equation.

$$x^{2}+6x+9-144=144-144$$

Subtracting 144 from itself leaves 0.

$$x^{2}+6x+9-144=0$$

Subtract 144 from 9.

$$x^{2}+6x-135=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and -135 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-135\right)}}{2}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-135\right)}}{2}$$

Multiply -4 times -135.

$$x=\frac{-6±\sqrt{36+540}}{2}$$

Add 36 to 540.

$$x=\frac{-6±\sqrt{576}}{2}$$

Take the square root of 576.

$$x=\frac{-6±24}{2}$$

Now solve the equation x=\frac{-6±24}{2} when ± is plus. Add -6 to 24.

$$x=\frac{18}{2}$$

Divide 18 by 2.

$$x=9$$

Now solve the equation x=\frac{-6±24}{2} when ± is minus. Subtract 24 from -6.

$$x=-\frac{30}{2}$$

Divide -30 by 2.

$$x=-15$$

The equation is now solved.

$$x=9 x=-15$$

Steps for Completing the Square

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+3\right)^{2}=144$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{144}$$

Simplify.

$$x+3=12 x+3=-12$$

Subtract 3 from both sides of the equation.

$$x=9 x=-15$$