Solve (u+3)^2=2u^2+8u-6

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

Solve for u

$$u=-5$$

$$u=3$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(u+3\right)^{2}.

$$u^{2}+6u+9=2u^{2}+8u-6$$

Subtract 2u^{2} from both sides.

$$u^{2}+6u+9-2u^{2}=8u-6$$

Combine u^{2} and -2u^{2} to get -u^{2}.

$$-u^{2}+6u+9=8u-6$$

Subtract 8u from both sides.

$$-u^{2}+6u+9-8u=-6$$

Combine 6u and -8u to get -2u.

$$-u^{2}-2u+9=-6$$

Add 6 to both sides.

$$-u^{2}-2u+9+6=0$$

Add 9 and 6 to get 15.

$$-u^{2}-2u+15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -u^{2}+au+bu+15. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-15=-15$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -15.

$$1,-15 3,-5$$

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -2.

$$a=3 b=-5$$

Rewrite -u^{2}-2u+15 as \left(-u^{2}+3u\right)+\left(-5u+15\right).

$$\left(-u^{2}+3u\right)+\left(-5u+15\right)$$

Factor out u in the first and 5 in the second group.

$$u\left(-u+3\right)+5\left(-u+3\right)$$

Factor out common term -u+3 by using distributive property.

$$\left(-u+3\right)\left(u+5\right)$$

To find equation solutions, solve -u+3=0 and u+5=0.

$$u=3 u=-5$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(u+3\right)^{2}.

$$u^{2}+6u+9=2u^{2}+8u-6$$

Subtract 2u^{2} from both sides.

$$u^{2}+6u+9-2u^{2}=8u-6$$

Combine u^{2} and -2u^{2} to get -u^{2}.

$$-u^{2}+6u+9=8u-6$$

Subtract 8u from both sides.

$$-u^{2}+6u+9-8u=-6$$

Combine 6u and -8u to get -2u.

$$-u^{2}-2u+9=-6$$

Add 6 to both sides.

$$-u^{2}-2u+9+6=0$$

Add 9 and 6 to get 15.

$$-u^{2}-2u+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -2 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$u=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-1\right)\times 15}}{2\left(-1\right)}$$

Square -2.

$$u=\frac{-\left(-2\right)±\sqrt{4-4\left(-1\right)\times 15}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$u=\frac{-\left(-2\right)±\sqrt{4+4\times 15}}{2\left(-1\right)}$$

Multiply 4 times 15.

$$u=\frac{-\left(-2\right)±\sqrt{4+60}}{2\left(-1\right)}$$

Add 4 to 60.

$$u=\frac{-\left(-2\right)±\sqrt{64}}{2\left(-1\right)}$$

Take the square root of 64.

$$u=\frac{-\left(-2\right)±8}{2\left(-1\right)}$$

The opposite of -2 is 2.

$$u=\frac{2±8}{2\left(-1\right)}$$

Multiply 2 times -1.

$$u=\frac{2±8}{-2}$$

Now solve the equation u=\frac{2±8}{-2} when ± is plus. Add 2 to 8.

$$u=\frac{10}{-2}$$

Divide 10 by -2.

$$u=-5$$

Now solve the equation u=\frac{2±8}{-2} when ± is minus. Subtract 8 from 2.

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

Divide -6 by -2.

$$u=3$$

The equation is now solved.

$$u=-5 u=3$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(u+3\right)^{2}.

$$u^{2}+6u+9=2u^{2}+8u-6$$

Subtract 2u^{2} from both sides.

$$u^{2}+6u+9-2u^{2}=8u-6$$

Combine u^{2} and -2u^{2} to get -u^{2}.

$$-u^{2}+6u+9=8u-6$$

Subtract 8u from both sides.

$$-u^{2}+6u+9-8u=-6$$

Combine 6u and -8u to get -2u.

$$-u^{2}-2u+9=-6$$

Subtract 9 from both sides.

$$-u^{2}-2u=-6-9$$

Subtract 9 from -6 to get -15.

$$-u^{2}-2u=-15$$

Divide both sides by -1.

$$\frac{-u^{2}-2u}{-1}=-\frac{15}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$u^{2}+\left(-\frac{2}{-1}\right)u=-\frac{15}{-1}$$

Divide -2 by -1.

$$u^{2}+2u=-\frac{15}{-1}$$

Divide -15 by -1.

$$u^{2}+2u=15$$

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

$$u^{2}+2u+1^{2}=15+1^{2}$$

Square 1.

$$u^{2}+2u+1=15+1$$

Add 15 to 1.

$$u^{2}+2u+1=16$$

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

$$\left(u+1\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(u+1\right)^{2}}=\sqrt{16}$$

Simplify.

$$u+1=4 u+1=-4$$

Subtract 1 from both sides of the equation.

$$u=3 u=-5$$