Solve 2u^2-5u+37=(u+3)^2

$$2 u ^ { 2 } - 5 u + 37 = ( u + 3 ) ^ { 2 }$$

Solve for u

$$u=4$$

$$u=7$$

Steps Using Factoring

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

$$2u^{2}-5u+37=u^{2}+6u+9$$

Subtract u^{2} from both sides.

$$2u^{2}-5u+37-u^{2}=6u+9$$

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

$$u^{2}-5u+37=6u+9$$

Subtract 6u from both sides.

$$u^{2}-5u+37-6u=9$$

Combine -5u and -6u to get -11u.

$$u^{2}-11u+37=9$$

Subtract 9 from both sides.

$$u^{2}-11u+37-9=0$$

Subtract 9 from 37 to get 28.

$$u^{2}-11u+28=0$$

To solve the equation, factor u^{2}-11u+28 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.

$$a+b=-11 ab=28$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 28.

$$-1,-28 -2,-14 -4,-7$$

Calculate the sum for each pair.

$$-1-28=-29 -2-14=-16 -4-7=-11$$

The solution is the pair that gives sum -11.

$$a=-7 b=-4$$

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

$$\left(u-7\right)\left(u-4\right)$$

To find equation solutions, solve u-7=0 and u-4=0.

$$u=7 u=4$$

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}.

$$2u^{2}-5u+37=u^{2}+6u+9$$

Subtract u^{2} from both sides.

$$2u^{2}-5u+37-u^{2}=6u+9$$

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

$$u^{2}-5u+37=6u+9$$

Subtract 6u from both sides.

$$u^{2}-5u+37-6u=9$$

Combine -5u and -6u to get -11u.

$$u^{2}-11u+37=9$$

Subtract 9 from both sides.

$$u^{2}-11u+37-9=0$$

Subtract 9 from 37 to get 28.

$$u^{2}-11u+28=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as u^{2}+au+bu+28. To find a and b, set up a system to be solved.

$$a+b=-11 ab=1\times 28=28$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 28.

$$-1,-28 -2,-14 -4,-7$$

Calculate the sum for each pair.

$$-1-28=-29 -2-14=-16 -4-7=-11$$

The solution is the pair that gives sum -11.

$$a=-7 b=-4$$

Rewrite u^{2}-11u+28 as \left(u^{2}-7u\right)+\left(-4u+28\right).

$$\left(u^{2}-7u\right)+\left(-4u+28\right)$$

Factor out u in the first and -4 in the second group.

$$u\left(u-7\right)-4\left(u-7\right)$$

Factor out common term u-7 by using distributive property.

$$\left(u-7\right)\left(u-4\right)$$

To find equation solutions, solve u-7=0 and u-4=0.

$$u=7 u=4$$

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}.

$$2u^{2}-5u+37=u^{2}+6u+9$$

Subtract u^{2} from both sides.

$$2u^{2}-5u+37-u^{2}=6u+9$$

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

$$u^{2}-5u+37=6u+9$$

Subtract 6u from both sides.

$$u^{2}-5u+37-6u=9$$

Combine -5u and -6u to get -11u.

$$u^{2}-11u+37=9$$

Subtract 9 from both sides.

$$u^{2}-11u+37-9=0$$

Subtract 9 from 37 to get 28.

$$u^{2}-11u+28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -11 for b, and 28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$u=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 28}}{2}$$

Square -11.

$$u=\frac{-\left(-11\right)±\sqrt{121-4\times 28}}{2}$$

Multiply -4 times 28.

$$u=\frac{-\left(-11\right)±\sqrt{121-112}}{2}$$

Add 121 to -112.

$$u=\frac{-\left(-11\right)±\sqrt{9}}{2}$$

Take the square root of 9.

$$u=\frac{-\left(-11\right)±3}{2}$$

The opposite of -11 is 11.

$$u=\frac{11±3}{2}$$

Now solve the equation u=\frac{11±3}{2} when ± is plus. Add 11 to 3.

$$u=\frac{14}{2}$$

Divide 14 by 2.

$$u=7$$

Now solve the equation u=\frac{11±3}{2} when ± is minus. Subtract 3 from 11.

$$u=\frac{8}{2}$$

Divide 8 by 2.

$$u=4$$

The equation is now solved.

$$u=7 u=4$$

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}.

$$2u^{2}-5u+37=u^{2}+6u+9$$

Subtract u^{2} from both sides.

$$2u^{2}-5u+37-u^{2}=6u+9$$

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

$$u^{2}-5u+37=6u+9$$

Subtract 6u from both sides.

$$u^{2}-5u+37-6u=9$$

Combine -5u and -6u to get -11u.

$$u^{2}-11u+37=9$$

Subtract 37 from both sides.

$$u^{2}-11u=9-37$$

Subtract 37 from 9 to get -28.

$$u^{2}-11u=-28$$

Divide -11, the coefficient of the x term, by 2 to get -\frac{11}{2}. Then add the square of -\frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$u^{2}-11u+\left(-\frac{11}{2}\right)^{2}=-28+\left(-\frac{11}{2}\right)^{2}$$

Square -\frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$u^{2}-11u+\frac{121}{4}=-28+\frac{121}{4}$$

Add -28 to \frac{121}{4}.

$$u^{2}-11u+\frac{121}{4}=\frac{9}{4}$$

Factor u^{2}-11u+\frac{121}{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}.

$$\left(u-\frac{11}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(u-\frac{11}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$u-\frac{11}{2}=\frac{3}{2} u-\frac{11}{2}=-\frac{3}{2}$$

Add \frac{11}{2} to both sides of the equation.

$$u=7 u=4$$