Solve 25+y^2-4y+4=9-y^2+8y-16

$$25 + y ^ { 2 } - 4 y + 4 = 9 - y ^ { 2 } + 8 y - 16$$

Solve for y

$$y=3+3i$$

$$y=3-3i$$

Steps Using the Quadratic Formula

Add 25 and 4 to get 29.

$$29+y^{2}-4y=9-y^{2}+8y-16$$

Subtract 16 from 9 to get -7.

$$29+y^{2}-4y=-7-y^{2}+8y$$

Subtract -7 from both sides.

$$29+y^{2}-4y-\left(-7\right)=-y^{2}+8y$$

The opposite of -7 is 7.

$$29+y^{2}-4y+7=-y^{2}+8y$$

Add y^{2} to both sides.

$$29+y^{2}-4y+7+y^{2}=8y$$

Add 29 and 7 to get 36.

$$36+y^{2}-4y+y^{2}=8y$$

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

$$36+2y^{2}-4y=8y$$

Subtract 8y from both sides.

$$36+2y^{2}-4y-8y=0$$

Combine -4y and -8y to get -12y.

$$36+2y^{2}-12y=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2y^{2}-12y+36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -12 for b, and 36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 2\times 36}}{2\times 2}$$

Square -12.

$$y=\frac{-\left(-12\right)±\sqrt{144-4\times 2\times 36}}{2\times 2}$$

Multiply -4 times 2.

$$y=\frac{-\left(-12\right)±\sqrt{144-8\times 36}}{2\times 2}$$

Multiply -8 times 36.

$$y=\frac{-\left(-12\right)±\sqrt{144-288}}{2\times 2}$$

Add 144 to -288.

$$y=\frac{-\left(-12\right)±\sqrt{-144}}{2\times 2}$$

Take the square root of -144.

$$y=\frac{-\left(-12\right)±12i}{2\times 2}$$

The opposite of -12 is 12.

$$y=\frac{12±12i}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{12±12i}{4}$$

Now solve the equation y=\frac{12±12i}{4} when ± is plus. Add 12 to 12i.

$$y=\frac{12+12i}{4}$$

Divide 12+12i by 4.

$$y=3+3i$$

Now solve the equation y=\frac{12±12i}{4} when ± is minus. Subtract 12i from 12.

$$y=\frac{12-12i}{4}$$

Divide 12-12i by 4.

$$y=3-3i$$

The equation is now solved.

$$y=3+3i y=3-3i$$

Steps for Completing the Square

Add 25 and 4 to get 29.

$$29+y^{2}-4y=9-y^{2}+8y-16$$

Subtract 16 from 9 to get -7.

$$29+y^{2}-4y=-7-y^{2}+8y$$

Add y^{2} to both sides.

$$29+y^{2}-4y+y^{2}=-7+8y$$

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

$$29+2y^{2}-4y=-7+8y$$

Subtract 8y from both sides.

$$29+2y^{2}-4y-8y=-7$$

Combine -4y and -8y to get -12y.

$$29+2y^{2}-12y=-7$$

Subtract 29 from both sides.

$$2y^{2}-12y=-7-29$$

Subtract 29 from -7 to get -36.

$$2y^{2}-12y=-36$$

Divide both sides by 2.

$$\frac{2y^{2}-12y}{2}=-\frac{36}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$y^{2}+\left(-\frac{12}{2}\right)y=-\frac{36}{2}$$

Divide -12 by 2.

$$y^{2}-6y=-\frac{36}{2}$$

Divide -36 by 2.

$$y^{2}-6y=-18$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-6y+\left(-3\right)^{2}=-18+\left(-3\right)^{2}$$

Square -3.

$$y^{2}-6y+9=-18+9$$

Add -18 to 9.

$$y^{2}-6y+9=-9$$

Factor y^{2}-6y+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-3\right)^{2}=-9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-3\right)^{2}}=\sqrt{-9}$$

Simplify.

$$y-3=3i y-3=-3i$$

Add 3 to both sides of the equation.

$$y=3+3i y=3-3i$$