Solve (y-5)^2=2y^2-15y+11

$$( y - 5 ) ^ { 2 } = 2 y ^ { 2 } - 15 y + 11$$

Solve for y

$$y=-2$$

$$y=7$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-5\right)^{2}.

$$y^{2}-10y+25=2y^{2}-15y+11$$

Subtract 2y^{2} from both sides.

$$y^{2}-10y+25-2y^{2}=-15y+11$$

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

$$-y^{2}-10y+25=-15y+11$$

Add 15y to both sides.

$$-y^{2}-10y+25+15y=11$$

Combine -10y and 15y to get 5y.

$$-y^{2}+5y+25=11$$

Subtract 11 from both sides.

$$-y^{2}+5y+25-11=0$$

Subtract 11 from 25 to get 14.

$$-y^{2}+5y+14=0$$

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

$$a+b=5 ab=-14=-14$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -14.

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

Calculate the sum for each pair.

$$-1+14=13 -2+7=5$$

The solution is the pair that gives sum 5.

$$a=7 b=-2$$

Rewrite -y^{2}+5y+14 as \left(-y^{2}+7y\right)+\left(-2y+14\right).

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

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

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

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

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

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

$$y=7 y=-2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-5\right)^{2}.

$$y^{2}-10y+25=2y^{2}-15y+11$$

Subtract 2y^{2} from both sides.

$$y^{2}-10y+25-2y^{2}=-15y+11$$

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

$$-y^{2}-10y+25=-15y+11$$

Add 15y to both sides.

$$-y^{2}-10y+25+15y=11$$

Combine -10y and 15y to get 5y.

$$-y^{2}+5y+25=11$$

Subtract 11 from both sides.

$$-y^{2}+5y+25-11=0$$

Subtract 11 from 25 to get 14.

$$-y^{2}+5y+14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 5 for b, and 14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\times 14}}{2\left(-1\right)}$$

Square 5.

$$y=\frac{-5±\sqrt{25-4\left(-1\right)\times 14}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$y=\frac{-5±\sqrt{25+4\times 14}}{2\left(-1\right)}$$

Multiply 4 times 14.

$$y=\frac{-5±\sqrt{25+56}}{2\left(-1\right)}$$

Add 25 to 56.

$$y=\frac{-5±\sqrt{81}}{2\left(-1\right)}$$

Take the square root of 81.

$$y=\frac{-5±9}{2\left(-1\right)}$$

Multiply 2 times -1.

$$y=\frac{-5±9}{-2}$$

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

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

Divide 4 by -2.

$$y=-2$$

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

$$y=-\frac{14}{-2}$$

Divide -14 by -2.

$$y=7$$

The equation is now solved.

$$y=-2 y=7$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-5\right)^{2}.

$$y^{2}-10y+25=2y^{2}-15y+11$$

Subtract 2y^{2} from both sides.

$$y^{2}-10y+25-2y^{2}=-15y+11$$

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

$$-y^{2}-10y+25=-15y+11$$

Add 15y to both sides.

$$-y^{2}-10y+25+15y=11$$

Combine -10y and 15y to get 5y.

$$-y^{2}+5y+25=11$$

Subtract 25 from both sides.

$$-y^{2}+5y=11-25$$

Subtract 25 from 11 to get -14.

$$-y^{2}+5y=-14$$

Divide both sides by -1.

$$\frac{-y^{2}+5y}{-1}=-\frac{14}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$y^{2}+\frac{5}{-1}y=-\frac{14}{-1}$$

Divide 5 by -1.

$$y^{2}-5y=-\frac{14}{-1}$$

Divide -14 by -1.

$$y^{2}-5y=14$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-5y+\left(-\frac{5}{2}\right)^{2}=14+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-5y+\frac{25}{4}=14+\frac{25}{4}$$

Add 14 to \frac{25}{4}.

$$y^{2}-5y+\frac{25}{4}=\frac{81}{4}$$

Factor y^{2}-5y+\frac{25}{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}.

$$\left(y-\frac{5}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{5}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

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

Add \frac{5}{2} to both sides of the equation.

$$y=7 y=-2$$