Solve 2u^2-21u+50=(u-6)^2

$$2 u ^ { 2 } - 21 u + 50 = ( u - 6 ) ^ { 2 }$$

Solve for u

$$u=2$$

$$u=7$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(u-6\right)^{2}.

$$2u^{2}-21u+50=u^{2}-12u+36$$

Subtract u^{2} from both sides.

$$2u^{2}-21u+50-u^{2}=-12u+36$$

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

$$u^{2}-21u+50=-12u+36$$

Add 12u to both sides.

$$u^{2}-21u+50+12u=36$$

Combine -21u and 12u to get -9u.

$$u^{2}-9u+50=36$$

Subtract 36 from both sides.

$$u^{2}-9u+50-36=0$$

Subtract 36 from 50 to get 14.

$$u^{2}-9u+14=0$$

To solve the equation, factor u^{2}-9u+14 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.

$$a+b=-9 ab=14$$

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 14.

$$-1,-14 -2,-7$$

Calculate the sum for each pair.

$$-1-14=-15 -2-7=-9$$

The solution is the pair that gives sum -9.

$$a=-7 b=-2$$

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

$$\left(u-7\right)\left(u-2\right)$$

To find equation solutions, solve u-7=0 and u-2=0.

$$u=7 u=2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(u-6\right)^{2}.

$$2u^{2}-21u+50=u^{2}-12u+36$$

Subtract u^{2} from both sides.

$$2u^{2}-21u+50-u^{2}=-12u+36$$

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

$$u^{2}-21u+50=-12u+36$$

Add 12u to both sides.

$$u^{2}-21u+50+12u=36$$

Combine -21u and 12u to get -9u.

$$u^{2}-9u+50=36$$

Subtract 36 from both sides.

$$u^{2}-9u+50-36=0$$

Subtract 36 from 50 to get 14.

$$u^{2}-9u+14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as u^{2}+au+bu+14. To find a and b, set up a system to be solved.

$$a+b=-9 ab=1\times 14=14$$

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 14.

$$-1,-14 -2,-7$$

Calculate the sum for each pair.

$$-1-14=-15 -2-7=-9$$

The solution is the pair that gives sum -9.

$$a=-7 b=-2$$

Rewrite u^{2}-9u+14 as \left(u^{2}-7u\right)+\left(-2u+14\right).

$$\left(u^{2}-7u\right)+\left(-2u+14\right)$$

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

$$u\left(u-7\right)-2\left(u-7\right)$$

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

$$\left(u-7\right)\left(u-2\right)$$

To find equation solutions, solve u-7=0 and u-2=0.

$$u=7 u=2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(u-6\right)^{2}.

$$2u^{2}-21u+50=u^{2}-12u+36$$

Subtract u^{2} from both sides.

$$2u^{2}-21u+50-u^{2}=-12u+36$$

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

$$u^{2}-21u+50=-12u+36$$

Add 12u to both sides.

$$u^{2}-21u+50+12u=36$$

Combine -21u and 12u to get -9u.

$$u^{2}-9u+50=36$$

Subtract 36 from both sides.

$$u^{2}-9u+50-36=0$$

Subtract 36 from 50 to get 14.

$$u^{2}-9u+14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -9 for b, and 14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$u=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 14}}{2}$$

Square -9.

$$u=\frac{-\left(-9\right)±\sqrt{81-4\times 14}}{2}$$

Multiply -4 times 14.

$$u=\frac{-\left(-9\right)±\sqrt{81-56}}{2}$$

Add 81 to -56.

$$u=\frac{-\left(-9\right)±\sqrt{25}}{2}$$

Take the square root of 25.

$$u=\frac{-\left(-9\right)±5}{2}$$

The opposite of -9 is 9.

$$u=\frac{9±5}{2}$$

Now solve the equation u=\frac{9±5}{2} when ± is plus. Add 9 to 5.

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

Divide 14 by 2.

$$u=7$$

Now solve the equation u=\frac{9±5}{2} when ± is minus. Subtract 5 from 9.

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

Divide 4 by 2.

$$u=2$$

The equation is now solved.

$$u=7 u=2$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(u-6\right)^{2}.

$$2u^{2}-21u+50=u^{2}-12u+36$$

Subtract u^{2} from both sides.

$$2u^{2}-21u+50-u^{2}=-12u+36$$

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

$$u^{2}-21u+50=-12u+36$$

Add 12u to both sides.

$$u^{2}-21u+50+12u=36$$

Combine -21u and 12u to get -9u.

$$u^{2}-9u+50=36$$

Subtract 50 from both sides.

$$u^{2}-9u=36-50$$

Subtract 50 from 36 to get -14.

$$u^{2}-9u=-14$$

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.

$$u^{2}-9u+\left(-\frac{9}{2}\right)^{2}=-14+\left(-\frac{9}{2}\right)^{2}$$

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

$$u^{2}-9u+\frac{81}{4}=-14+\frac{81}{4}$$

Add -14 to \frac{81}{4}.

$$u^{2}-9u+\frac{81}{4}=\frac{25}{4}$$

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}.

$$\left(u-\frac{9}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(u-\frac{9}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$u-\frac{9}{2}=\frac{5}{2} u-\frac{9}{2}=-\frac{5}{2}$$

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

$$u=7 u=2$$