Solve (y-2)^2=-y^2+10

$$( y - 2 ) ^ { 2 } = - y ^ { 2 } + 10$$

Solve for y

$$y=-1$$

$$y=3$$

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}.

$$y^{2}-4y+4=-y^{2}+10$$

Add y^{2} to both sides.

$$y^{2}-4y+4+y^{2}=10$$

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

$$2y^{2}-4y+4=10$$

Subtract 10 from both sides.

$$2y^{2}-4y+4-10=0$$

Subtract 10 from 4 to get -6.

$$2y^{2}-4y-6=0$$

Divide both sides by 2.

$$y^{2}-2y-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\left(-3\right)=-3$$

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.

$$a=-3 b=1$$

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

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

Factor out y in y^{2}-3y.

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

Factor out common term y-3 by using distributive property.

$$\left(y-3\right)\left(y+1\right)$$

To find equation solutions, solve y-3=0 and y+1=0.

$$y=3 y=-1$$

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}.

$$y^{2}-4y+4=-y^{2}+10$$

Add y^{2} to both sides.

$$y^{2}-4y+4+y^{2}=10$$

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

$$2y^{2}-4y+4=10$$

Subtract 10 from both sides.

$$2y^{2}-4y+4-10=0$$

Subtract 10 from 4 to get -6.

$$2y^{2}-4y-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -4 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 2\left(-6\right)}}{2\times 2}$$

Square -4.

$$y=\frac{-\left(-4\right)±\sqrt{16-4\times 2\left(-6\right)}}{2\times 2}$$

Multiply -4 times 2.

$$y=\frac{-\left(-4\right)±\sqrt{16-8\left(-6\right)}}{2\times 2}$$

Multiply -8 times -6.

$$y=\frac{-\left(-4\right)±\sqrt{16+48}}{2\times 2}$$

Add 16 to 48.

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

Take the square root of 64.

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

The opposite of -4 is 4.

$$y=\frac{4±8}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{4±8}{4}$$

Now solve the equation y=\frac{4±8}{4} when ± is plus. Add 4 to 8.

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

Divide 12 by 4.

$$y=3$$

Now solve the equation y=\frac{4±8}{4} when ± is minus. Subtract 8 from 4.

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

Divide -4 by 4.

$$y=-1$$

The equation is now solved.

$$y=3 y=-1$$

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}.

$$y^{2}-4y+4=-y^{2}+10$$

Add y^{2} to both sides.

$$y^{2}-4y+4+y^{2}=10$$

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

$$2y^{2}-4y+4=10$$

Subtract 4 from both sides.

$$2y^{2}-4y=10-4$$

Subtract 4 from 10 to get 6.

$$2y^{2}-4y=6$$

Divide both sides by 2.

$$\frac{2y^{2}-4y}{2}=\frac{6}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$y^{2}+\left(-\frac{4}{2}\right)y=\frac{6}{2}$$

Divide -4 by 2.

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

Divide 6 by 2.

$$y^{2}-2y=3$$

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-2y+1=3+1$$

Add 3 to 1.

$$y^{2}-2y+1=4$$

Factor y^{2}-2y+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$y-1=2 y-1=-2$$

Add 1 to both sides of the equation.

$$y=3 y=-1$$