Solve 2u^2+4u-2=(u+1)^2

$$2 u ^ { 2 } + 4 u - 2 = ( u + 1 ) ^ { 2 }$$

Solve for u

$$u=-3$$

$$u=1$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(u+1\right)^{2}.

$$2u^{2}+4u-2=u^{2}+2u+1$$

Subtract u^{2} from both sides.

$$2u^{2}+4u-2-u^{2}=2u+1$$

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

$$u^{2}+4u-2=2u+1$$

Subtract 2u from both sides.

$$u^{2}+4u-2-2u=1$$

Combine 4u and -2u to get 2u.

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

Subtract 1 from both sides.

$$u^{2}+2u-2-1=0$$

Subtract 1 from -2 to get -3.

$$u^{2}+2u-3=0$$

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

$$a+b=2 ab=-3$$

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. The only such pair is the system solution.

$$a=-1 b=3$$

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

$$\left(u-1\right)\left(u+3\right)$$

To find equation solutions, solve u-1=0 and u+3=0.

$$u=1 u=-3$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(u+1\right)^{2}.

$$2u^{2}+4u-2=u^{2}+2u+1$$

Subtract u^{2} from both sides.

$$2u^{2}+4u-2-u^{2}=2u+1$$

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

$$u^{2}+4u-2=2u+1$$

Subtract 2u from both sides.

$$u^{2}+4u-2-2u=1$$

Combine 4u and -2u to get 2u.

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

Subtract 1 from both sides.

$$u^{2}+2u-2-1=0$$

Subtract 1 from -2 to get -3.

$$u^{2}+2u-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as u^{2}+au+bu-3. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-3\right)=-3$$

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. The only such pair is the system solution.

$$a=-1 b=3$$

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

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

Factor out u in the first and 3 in the second group.

$$u\left(u-1\right)+3\left(u-1\right)$$

Factor out common term u-1 by using distributive property.

$$\left(u-1\right)\left(u+3\right)$$

To find equation solutions, solve u-1=0 and u+3=0.

$$u=1 u=-3$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(u+1\right)^{2}.

$$2u^{2}+4u-2=u^{2}+2u+1$$

Subtract u^{2} from both sides.

$$2u^{2}+4u-2-u^{2}=2u+1$$

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

$$u^{2}+4u-2=2u+1$$

Subtract 2u from both sides.

$$u^{2}+4u-2-2u=1$$

Combine 4u and -2u to get 2u.

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

Subtract 1 from both sides.

$$u^{2}+2u-2-1=0$$

Subtract 1 from -2 to get -3.

$$u^{2}+2u-3=0$$

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

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

Square 2.

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

Multiply -4 times -3.

$$u=\frac{-2±\sqrt{4+12}}{2}$$

Add 4 to 12.

$$u=\frac{-2±\sqrt{16}}{2}$$

Take the square root of 16.

$$u=\frac{-2±4}{2}$$

Now solve the equation u=\frac{-2±4}{2} when ± is plus. Add -2 to 4.

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

Divide 2 by 2.

$$u=1$$

Now solve the equation u=\frac{-2±4}{2} when ± is minus. Subtract 4 from -2.

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

Divide -6 by 2.

$$u=-3$$

The equation is now solved.

$$u=1 u=-3$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(u+1\right)^{2}.

$$2u^{2}+4u-2=u^{2}+2u+1$$

Subtract u^{2} from both sides.

$$2u^{2}+4u-2-u^{2}=2u+1$$

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

$$u^{2}+4u-2=2u+1$$

Subtract 2u from both sides.

$$u^{2}+4u-2-2u=1$$

Combine 4u and -2u to get 2u.

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

Add 2 to both sides.

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

Add 1 and 2 to get 3.

$$u^{2}+2u=3$$

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.

$$u^{2}+2u+1^{2}=3+1^{2}$$

Square 1.

$$u^{2}+2u+1=3+1$$

Add 3 to 1.

$$u^{2}+2u+1=4$$

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}.

$$\left(u+1\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(u+1\right)^{2}}=\sqrt{4}$$

Simplify.

$$u+1=2 u+1=-2$$

Subtract 1 from both sides of the equation.

$$u=1 u=-3$$