Solve (u-3)^2=2u^2-5u-21

$$( u - 3 ) ^ { 2 } = 2 u ^ { 2 } - 5 u - 21$$

Solve for u

$$u=-6$$

$$u=5$$

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}.

$$u^{2}-6u+9=2u^{2}-5u-21$$

Subtract 2u^{2} from both sides.

$$u^{2}-6u+9-2u^{2}=-5u-21$$

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

$$-u^{2}-6u+9=-5u-21$$

Add 5u to both sides.

$$-u^{2}-6u+9+5u=-21$$

Combine -6u and 5u to get -u.

$$-u^{2}-u+9=-21$$

Add 21 to both sides.

$$-u^{2}-u+9+21=0$$

Add 9 and 21 to get 30.

$$-u^{2}-u+30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -u^{2}+au+bu+30. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-30=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -1.

$$a=5 b=-6$$

Rewrite -u^{2}-u+30 as \left(-u^{2}+5u\right)+\left(-6u+30\right).

$$\left(-u^{2}+5u\right)+\left(-6u+30\right)$$

Factor out u in the first and 6 in the second group.

$$u\left(-u+5\right)+6\left(-u+5\right)$$

Factor out common term -u+5 by using distributive property.

$$\left(-u+5\right)\left(u+6\right)$$

To find equation solutions, solve -u+5=0 and u+6=0.

$$u=5 u=-6$$

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}.

$$u^{2}-6u+9=2u^{2}-5u-21$$

Subtract 2u^{2} from both sides.

$$u^{2}-6u+9-2u^{2}=-5u-21$$

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

$$-u^{2}-6u+9=-5u-21$$

Add 5u to both sides.

$$-u^{2}-6u+9+5u=-21$$

Combine -6u and 5u to get -u.

$$-u^{2}-u+9=-21$$

Add 21 to both sides.

$$-u^{2}-u+9+21=0$$

Add 9 and 21 to get 30.

$$-u^{2}-u+30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -1 for b, and 30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$u=\frac{-\left(-1\right)±\sqrt{1-4\left(-1\right)\times 30}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$u=\frac{-\left(-1\right)±\sqrt{1+4\times 30}}{2\left(-1\right)}$$

Multiply 4 times 30.

$$u=\frac{-\left(-1\right)±\sqrt{1+120}}{2\left(-1\right)}$$

Add 1 to 120.

$$u=\frac{-\left(-1\right)±\sqrt{121}}{2\left(-1\right)}$$

Take the square root of 121.

$$u=\frac{-\left(-1\right)±11}{2\left(-1\right)}$$

The opposite of -1 is 1.

$$u=\frac{1±11}{2\left(-1\right)}$$

Multiply 2 times -1.

$$u=\frac{1±11}{-2}$$

Now solve the equation u=\frac{1±11}{-2} when ± is plus. Add 1 to 11.

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

Divide 12 by -2.

$$u=-6$$

Now solve the equation u=\frac{1±11}{-2} when ± is minus. Subtract 11 from 1.

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

Divide -10 by -2.

$$u=5$$

The equation is now solved.

$$u=-6 u=5$$

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}.

$$u^{2}-6u+9=2u^{2}-5u-21$$

Subtract 2u^{2} from both sides.

$$u^{2}-6u+9-2u^{2}=-5u-21$$

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

$$-u^{2}-6u+9=-5u-21$$

Add 5u to both sides.

$$-u^{2}-6u+9+5u=-21$$

Combine -6u and 5u to get -u.

$$-u^{2}-u+9=-21$$

Subtract 9 from both sides.

$$-u^{2}-u=-21-9$$

Subtract 9 from -21 to get -30.

$$-u^{2}-u=-30$$

Divide both sides by -1.

$$\frac{-u^{2}-u}{-1}=-\frac{30}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$u^{2}+\left(-\frac{1}{-1}\right)u=-\frac{30}{-1}$$

Divide -1 by -1.

$$u^{2}+u=-\frac{30}{-1}$$

Divide -30 by -1.

$$u^{2}+u=30$$

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

$$u^{2}+u+\left(\frac{1}{2}\right)^{2}=30+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$u^{2}+u+\frac{1}{4}=30+\frac{1}{4}$$

Add 30 to \frac{1}{4}.

$$u^{2}+u+\frac{1}{4}=\frac{121}{4}$$

Factor u^{2}+u+\frac{1}{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}.

$$\left(u+\frac{1}{2}\right)^{2}=\frac{121}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(u+\frac{1}{2}\right)^{2}}=\sqrt{\frac{121}{4}}$$

Simplify.

$$u+\frac{1}{2}=\frac{11}{2} u+\frac{1}{2}=-\frac{11}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$u=5 u=-6$$