Solve (y-10)^2-13=12

$$( y - 10 ) ^ { 2 } - 13 = 12$$

Solve for y

$$y=15$$

$$y=5$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-10\right)^{2}.

$$y^{2}-20y+100-13=12$$

Subtract 13 from 100 to get 87.

$$y^{2}-20y+87=12$$

Subtract 12 from both sides.

$$y^{2}-20y+87-12=0$$

Subtract 12 from 87 to get 75.

$$y^{2}-20y+75=0$$

To solve the equation, factor y^{2}-20y+75 using formula y^{2}+\left(a+b\right)y+ab=\left(y+a\right)\left(y+b\right). To find a and b, set up a system to be solved.

$$a+b=-20 ab=75$$

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 75.

$$-1,-75 -3,-25 -5,-15$$

Calculate the sum for each pair.

$$-1-75=-76 -3-25=-28 -5-15=-20$$

The solution is the pair that gives sum -20.

$$a=-15 b=-5$$

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

$$\left(y-15\right)\left(y-5\right)$$

To find equation solutions, solve y-15=0 and y-5=0.

$$y=15 y=5$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-10\right)^{2}.

$$y^{2}-20y+100-13=12$$

Subtract 13 from 100 to get 87.

$$y^{2}-20y+87=12$$

Subtract 12 from both sides.

$$y^{2}-20y+87-12=0$$

Subtract 12 from 87 to get 75.

$$y^{2}-20y+75=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+75. To find a and b, set up a system to be solved.

$$a+b=-20 ab=1\times 75=75$$

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 75.

$$-1,-75 -3,-25 -5,-15$$

Calculate the sum for each pair.

$$-1-75=-76 -3-25=-28 -5-15=-20$$

The solution is the pair that gives sum -20.

$$a=-15 b=-5$$

Rewrite y^{2}-20y+75 as \left(y^{2}-15y\right)+\left(-5y+75\right).

$$\left(y^{2}-15y\right)+\left(-5y+75\right)$$

Factor out y in the first and -5 in the second group.

$$y\left(y-15\right)-5\left(y-15\right)$$

Factor out common term y-15 by using distributive property.

$$\left(y-15\right)\left(y-5\right)$$

To find equation solutions, solve y-15=0 and y-5=0.

$$y=15 y=5$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-10\right)^{2}.

$$y^{2}-20y+100-13=12$$

Subtract 13 from 100 to get 87.

$$y^{2}-20y+87=12$$

Subtract 12 from both sides.

$$y^{2}-20y+87-12=0$$

Subtract 12 from 87 to get 75.

$$y^{2}-20y+75=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -20 for b, and 75 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 75}}{2}$$

Square -20.

$$y=\frac{-\left(-20\right)±\sqrt{400-4\times 75}}{2}$$

Multiply -4 times 75.

$$y=\frac{-\left(-20\right)±\sqrt{400-300}}{2}$$

Add 400 to -300.

$$y=\frac{-\left(-20\right)±\sqrt{100}}{2}$$

Take the square root of 100.

$$y=\frac{-\left(-20\right)±10}{2}$$

The opposite of -20 is 20.

$$y=\frac{20±10}{2}$$

Now solve the equation y=\frac{20±10}{2} when ± is plus. Add 20 to 10.

$$y=\frac{30}{2}$$

Divide 30 by 2.

$$y=15$$

Now solve the equation y=\frac{20±10}{2} when ± is minus. Subtract 10 from 20.

$$y=\frac{10}{2}$$

Divide 10 by 2.

$$y=5$$

The equation is now solved.

$$y=15 y=5$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(y-10\right)^{2}.

$$y^{2}-20y+100-13=12$$

Subtract 13 from 100 to get 87.

$$y^{2}-20y+87=12$$

Subtract 87 from both sides.

$$y^{2}-20y=12-87$$

Subtract 87 from 12 to get -75.

$$y^{2}-20y=-75$$

Divide -20, the coefficient of the x term, by 2 to get -10. Then add the square of -10 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-20y+\left(-10\right)^{2}=-75+\left(-10\right)^{2}$$

Square -10.

$$y^{2}-20y+100=-75+100$$

Add -75 to 100.

$$y^{2}-20y+100=25$$

Factor y^{2}-20y+100. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-10\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-10\right)^{2}}=\sqrt{25}$$

Simplify.

$$y-10=5 y-10=-5$$

Add 10 to both sides of the equation.

$$y=15 y=5$$