Solve 5(y-2)^2-8=37

$$5 ( y - 2 ) ^ { 2 } - 8 = 37$$

Solve for y

$$y=5$$

$$y=-1$$

Steps Using Factoring By Grouping

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

$$5\left(y^{2}-4y+4\right)-8=37$$

Use the distributive property to multiply 5 by y^{2}-4y+4.

$$5y^{2}-20y+20-8=37$$

Subtract 8 from 20 to get 12.

$$5y^{2}-20y+12=37$$

Subtract 37 from both sides.

$$5y^{2}-20y+12-37=0$$

Subtract 37 from 12 to get -25.

$$5y^{2}-20y-25=0$$

Divide both sides by 5.

$$y^{2}-4y-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-5. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\left(-5\right)=-5$$

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. The only such pair is the system solution.

$$a=-5 b=1$$

Rewrite y^{2}-4y-5 as \left(y^{2}-5y\right)+\left(y-5\right).

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

Factor out y in y^{2}-5y.

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

Factor out common term y-5 by using distributive property.

$$\left(y-5\right)\left(y+1\right)$$

To find equation solutions, solve y-5=0 and y+1=0.

$$y=5 y=-1$$

Steps Using the Quadratic Formula

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

$$5\left(y^{2}-4y+4\right)-8=37$$

Use the distributive property to multiply 5 by y^{2}-4y+4.

$$5y^{2}-20y+20-8=37$$

Subtract 8 from 20 to get 12.

$$5y^{2}-20y+12=37$$

Subtract 37 from both sides.

$$5y^{2}-20y+12-37=0$$

Subtract 37 from 12 to get -25.

$$5y^{2}-20y-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -20 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 5\left(-25\right)}}{2\times 5}$$

Square -20.

$$y=\frac{-\left(-20\right)±\sqrt{400-4\times 5\left(-25\right)}}{2\times 5}$$

Multiply -4 times 5.

$$y=\frac{-\left(-20\right)±\sqrt{400-20\left(-25\right)}}{2\times 5}$$

Multiply -20 times -25.

$$y=\frac{-\left(-20\right)±\sqrt{400+500}}{2\times 5}$$

Add 400 to 500.

$$y=\frac{-\left(-20\right)±\sqrt{900}}{2\times 5}$$

Take the square root of 900.

$$y=\frac{-\left(-20\right)±30}{2\times 5}$$

The opposite of -20 is 20.

$$y=\frac{20±30}{2\times 5}$$

Multiply 2 times 5.

$$y=\frac{20±30}{10}$$

Now solve the equation y=\frac{20±30}{10} when ± is plus. Add 20 to 30.

$$y=\frac{50}{10}$$

Divide 50 by 10.

$$y=5$$

Now solve the equation y=\frac{20±30}{10} when ± is minus. Subtract 30 from 20.

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

Divide -10 by 10.

$$y=-1$$

The equation is now solved.

$$y=5 y=-1$$

Steps for Completing the Square

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

$$5\left(y^{2}-4y+4\right)-8=37$$

Use the distributive property to multiply 5 by y^{2}-4y+4.

$$5y^{2}-20y+20-8=37$$

Subtract 8 from 20 to get 12.

$$5y^{2}-20y+12=37$$

Subtract 12 from both sides.

$$5y^{2}-20y=37-12$$

Subtract 12 from 37 to get 25.

$$5y^{2}-20y=25$$

Divide both sides by 5.

$$\frac{5y^{2}-20y}{5}=\frac{25}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$y^{2}+\left(-\frac{20}{5}\right)y=\frac{25}{5}$$

Divide -20 by 5.

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

Divide 25 by 5.

$$y^{2}-4y=5$$

Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -2.

$$y^{2}-4y+4=5+4$$

Add 5 to 4.

$$y^{2}-4y+4=9$$

Factor y^{2}-4y+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}.

$$\left(y-2\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-2\right)^{2}}=\sqrt{9}$$

Simplify.

$$y-2=3 y-2=-3$$

Add 2 to both sides of the equation.

$$y=5 y=-1$$