Solve left(u+1right)^2=2u^2+5u+3

$${ \left( u+1 \right) }^{ 2 } =2 { u }^{ 2 } +5u+3$$

Solve for u

$$u=-1$$

$$u=-2$$

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}.

$$u^{2}+2u+1=2u^{2}+5u+3$$

Subtract 2u^{2} from both sides.

$$u^{2}+2u+1-2u^{2}=5u+3$$

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

$$-u^{2}+2u+1=5u+3$$

Subtract 5u from both sides.

$$-u^{2}+2u+1-5u=3$$

Combine 2u and -5u to get -3u.

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

Subtract 3 from both sides.

$$-u^{2}-3u+1-3=0$$

Subtract 3 from 1 to get -2.

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

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

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

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-1 b=-2$$

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

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

Factor out u in the first and 2 in the second group.

$$u\left(-u-1\right)+2\left(-u-1\right)$$

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

$$\left(-u-1\right)\left(u+2\right)$$

To find equation solutions, solve -u-1=0 and u+2=0.

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

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}.

$$u^{2}+2u+1=2u^{2}+5u+3$$

Subtract 2u^{2} from both sides.

$$u^{2}+2u+1-2u^{2}=5u+3$$

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

$$-u^{2}+2u+1=5u+3$$

Subtract 5u from both sides.

$$-u^{2}+2u+1-5u=3$$

Combine 2u and -5u to get -3u.

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

Subtract 3 from both sides.

$$-u^{2}-3u+1-3=0$$

Subtract 3 from 1 to get -2.

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

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

$$u=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)\left(-2\right)}}{2\left(-1\right)}$$

Square -3.

$$u=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\left(-2\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$u=\frac{-\left(-3\right)±\sqrt{9+4\left(-2\right)}}{2\left(-1\right)}$$

Multiply 4 times -2.

$$u=\frac{-\left(-3\right)±\sqrt{9-8}}{2\left(-1\right)}$$

Add 9 to -8.

$$u=\frac{-\left(-3\right)±\sqrt{1}}{2\left(-1\right)}$$

Take the square root of 1.

$$u=\frac{-\left(-3\right)±1}{2\left(-1\right)}$$

The opposite of -3 is 3.

$$u=\frac{3±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$u=\frac{3±1}{-2}$$

Now solve the equation u=\frac{3±1}{-2} when ± is plus. Add 3 to 1.

$$u=\frac{4}{-2}$$

Divide 4 by -2.

$$u=-2$$

Now solve the equation u=\frac{3±1}{-2} when ± is minus. Subtract 1 from 3.

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

Divide 2 by -2.

$$u=-1$$

The equation is now solved.

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

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}.

$$u^{2}+2u+1=2u^{2}+5u+3$$

Subtract 2u^{2} from both sides.

$$u^{2}+2u+1-2u^{2}=5u+3$$

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

$$-u^{2}+2u+1=5u+3$$

Subtract 5u from both sides.

$$-u^{2}+2u+1-5u=3$$

Combine 2u and -5u to get -3u.

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

Subtract 1 from both sides.

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

Subtract 1 from 3 to get 2.

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

Divide both sides by -1.

$$\frac{-u^{2}-3u}{-1}=\frac{2}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$u^{2}+\left(-\frac{3}{-1}\right)u=\frac{2}{-1}$$

Divide -3 by -1.

$$u^{2}+3u=\frac{2}{-1}$$

Divide 2 by -1.

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

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.

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

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

$$u^{2}+3u+\frac{9}{4}=-2+\frac{9}{4}$$

Add -2 to \frac{9}{4}.

$$u^{2}+3u+\frac{9}{4}=\frac{1}{4}$$

Factor u^{2}+3u+\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}.

$$\left(u+\frac{3}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(u+\frac{3}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$u+\frac{3}{2}=\frac{1}{2} u+\frac{3}{2}=-\frac{1}{2}$$

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

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