Solve (u+2)^2=2u^2-5u+22

$$( u + 2 ) ^ { 2 } = 2 u ^ { 2 } - 5 u + 22$$

Solve for u

$$u=3$$

$$u=6$$

Steps Using Factoring By Grouping

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

$$u^{2}+4u+4=2u^{2}-5u+22$$

Subtract 2u^{2} from both sides.

$$u^{2}+4u+4-2u^{2}=-5u+22$$

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

$$-u^{2}+4u+4=-5u+22$$

Add 5u to both sides.

$$-u^{2}+4u+4+5u=22$$

Combine 4u and 5u to get 9u.

$$-u^{2}+9u+4=22$$

Subtract 22 from both sides.

$$-u^{2}+9u+4-22=0$$

Subtract 22 from 4 to get -18.

$$-u^{2}+9u-18=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -u^{2}+au+bu-18. To find a and b, set up a system to be solved.

$$a+b=9 ab=-\left(-18\right)=18$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 18.

$$1,18 2,9 3,6$$

Calculate the sum for each pair.

$$1+18=19 2+9=11 3+6=9$$

The solution is the pair that gives sum 9.

$$a=6 b=3$$

Rewrite -u^{2}+9u-18 as \left(-u^{2}+6u\right)+\left(3u-18\right).

$$\left(-u^{2}+6u\right)+\left(3u-18\right)$$

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

$$-u\left(u-6\right)+3\left(u-6\right)$$

Factor out common term u-6 by using distributive property.

$$\left(u-6\right)\left(-u+3\right)$$

To find equation solutions, solve u-6=0 and -u+3=0.

$$u=6 u=3$$

Steps Using the Quadratic Formula

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

$$u^{2}+4u+4=2u^{2}-5u+22$$

Subtract 2u^{2} from both sides.

$$u^{2}+4u+4-2u^{2}=-5u+22$$

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

$$-u^{2}+4u+4=-5u+22$$

Add 5u to both sides.

$$-u^{2}+4u+4+5u=22$$

Combine 4u and 5u to get 9u.

$$-u^{2}+9u+4=22$$

Subtract 22 from both sides.

$$-u^{2}+9u+4-22=0$$

Subtract 22 from 4 to get -18.

$$-u^{2}+9u-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 9 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$u=\frac{-9±\sqrt{9^{2}-4\left(-1\right)\left(-18\right)}}{2\left(-1\right)}$$

Square 9.

$$u=\frac{-9±\sqrt{81-4\left(-1\right)\left(-18\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$u=\frac{-9±\sqrt{81+4\left(-18\right)}}{2\left(-1\right)}$$

Multiply 4 times -18.

$$u=\frac{-9±\sqrt{81-72}}{2\left(-1\right)}$$

Add 81 to -72.

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

Take the square root of 9.

$$u=\frac{-9±3}{2\left(-1\right)}$$

Multiply 2 times -1.

$$u=\frac{-9±3}{-2}$$

Now solve the equation u=\frac{-9±3}{-2} when ± is plus. Add -9 to 3.

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

Divide -6 by -2.

$$u=3$$

Now solve the equation u=\frac{-9±3}{-2} when ± is minus. Subtract 3 from -9.

$$u=-\frac{12}{-2}$$

Divide -12 by -2.

$$u=6$$

The equation is now solved.

$$u=3 u=6$$

Steps for Completing the Square

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

$$u^{2}+4u+4=2u^{2}-5u+22$$

Subtract 2u^{2} from both sides.

$$u^{2}+4u+4-2u^{2}=-5u+22$$

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

$$-u^{2}+4u+4=-5u+22$$

Add 5u to both sides.

$$-u^{2}+4u+4+5u=22$$

Combine 4u and 5u to get 9u.

$$-u^{2}+9u+4=22$$

Subtract 4 from both sides.

$$-u^{2}+9u=22-4$$

Subtract 4 from 22 to get 18.

$$-u^{2}+9u=18$$

Divide both sides by -1.

$$\frac{-u^{2}+9u}{-1}=\frac{18}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$u^{2}+\frac{9}{-1}u=\frac{18}{-1}$$

Divide 9 by -1.

$$u^{2}-9u=\frac{18}{-1}$$

Divide 18 by -1.

$$u^{2}-9u=-18$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$u^{2}-9u+\left(-\frac{9}{2}\right)^{2}=-18+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$u^{2}-9u+\frac{81}{4}=-18+\frac{81}{4}$$

Add -18 to \frac{81}{4}.

$$u^{2}-9u+\frac{81}{4}=\frac{9}{4}$$

Factor u^{2}-9u+\frac{81}{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}.

$$\left(u-\frac{9}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

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

Simplify.

$$u-\frac{9}{2}=\frac{3}{2} u-\frac{9}{2}=-\frac{3}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$u=6 u=3$$