Solve (y+2)^2=2y^2-4y+19

$$( y + 2 ) ^ { 2 } = 2 y ^ { 2 } - 4 y + 19$$

Solve for y

$$y=3$$

$$y=5$$

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}.

$$y^{2}+4y+4=2y^{2}-4y+19$$

Subtract 2y^{2} from both sides.

$$y^{2}+4y+4-2y^{2}=-4y+19$$

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

$$-y^{2}+4y+4=-4y+19$$

Add 4y to both sides.

$$-y^{2}+4y+4+4y=19$$

Combine 4y and 4y to get 8y.

$$-y^{2}+8y+4=19$$

Subtract 19 from both sides.

$$-y^{2}+8y+4-19=0$$

Subtract 19 from 4 to get -15.

$$-y^{2}+8y-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -y^{2}+ay+by-15. To find a and b, set up a system to be solved.

$$a+b=8 ab=-\left(-15\right)=15$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 15.

$$1,15 3,5$$

Calculate the sum for each pair.

$$1+15=16 3+5=8$$

The solution is the pair that gives sum 8.

$$a=5 b=3$$

Rewrite -y^{2}+8y-15 as \left(-y^{2}+5y\right)+\left(3y-15\right).

$$\left(-y^{2}+5y\right)+\left(3y-15\right)$$

Factor out -y in the first and 3 in the second group.

$$-y\left(y-5\right)+3\left(y-5\right)$$

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

$$\left(y-5\right)\left(-y+3\right)$$

To find equation solutions, solve y-5=0 and -y+3=0.

$$y=5 y=3$$

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}.

$$y^{2}+4y+4=2y^{2}-4y+19$$

Subtract 2y^{2} from both sides.

$$y^{2}+4y+4-2y^{2}=-4y+19$$

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

$$-y^{2}+4y+4=-4y+19$$

Add 4y to both sides.

$$-y^{2}+4y+4+4y=19$$

Combine 4y and 4y to get 8y.

$$-y^{2}+8y+4=19$$

Subtract 19 from both sides.

$$-y^{2}+8y+4-19=0$$

Subtract 19 from 4 to get -15.

$$-y^{2}+8y-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 8 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-8±\sqrt{8^{2}-4\left(-1\right)\left(-15\right)}}{2\left(-1\right)}$$

Square 8.

$$y=\frac{-8±\sqrt{64-4\left(-1\right)\left(-15\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$y=\frac{-8±\sqrt{64+4\left(-15\right)}}{2\left(-1\right)}$$

Multiply 4 times -15.

$$y=\frac{-8±\sqrt{64-60}}{2\left(-1\right)}$$

Add 64 to -60.

$$y=\frac{-8±\sqrt{4}}{2\left(-1\right)}$$

Take the square root of 4.

$$y=\frac{-8±2}{2\left(-1\right)}$$

Multiply 2 times -1.

$$y=\frac{-8±2}{-2}$$

Now solve the equation y=\frac{-8±2}{-2} when ± is plus. Add -8 to 2.

$$y=-\frac{6}{-2}$$

Divide -6 by -2.

$$y=3$$

Now solve the equation y=\frac{-8±2}{-2} when ± is minus. Subtract 2 from -8.

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

Divide -10 by -2.

$$y=5$$

The equation is now solved.

$$y=3 y=5$$

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}.

$$y^{2}+4y+4=2y^{2}-4y+19$$

Subtract 2y^{2} from both sides.

$$y^{2}+4y+4-2y^{2}=-4y+19$$

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

$$-y^{2}+4y+4=-4y+19$$

Add 4y to both sides.

$$-y^{2}+4y+4+4y=19$$

Combine 4y and 4y to get 8y.

$$-y^{2}+8y+4=19$$

Subtract 4 from both sides.

$$-y^{2}+8y=19-4$$

Subtract 4 from 19 to get 15.

$$-y^{2}+8y=15$$

Divide both sides by -1.

$$\frac{-y^{2}+8y}{-1}=\frac{15}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$y^{2}+\frac{8}{-1}y=\frac{15}{-1}$$

Divide 8 by -1.

$$y^{2}-8y=\frac{15}{-1}$$

Divide 15 by -1.

$$y^{2}-8y=-15$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-8y+\left(-4\right)^{2}=-15+\left(-4\right)^{2}$$

Square -4.

$$y^{2}-8y+16=-15+16$$

Add -15 to 16.

$$y^{2}-8y+16=1$$

Factor y^{2}-8y+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-4\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-4\right)^{2}}=\sqrt{1}$$

Simplify.

$$y-4=1 y-4=-1$$

Add 4 to both sides of the equation.

$$y=5 y=3$$