Solve (y+20)^2+y^2+20y+y^2=101

$$( y + 20 ) ^ { 2 } + y ^ { 2 } + 20 y + y ^ { 2 } = 101$$

Solve for y

$$y=\frac{\sqrt{3}}{3}-10\approx -9.422649731$$

$$y=-\frac{\sqrt{3}}{3}-10\approx -10.577350269$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+20\right)^{2}.

$$y^{2}+40y+400+y^{2}+20y+y^{2}=101$$

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

$$2y^{2}+40y+400+20y+y^{2}=101$$

Combine 40y and 20y to get 60y.

$$2y^{2}+60y+400+y^{2}=101$$

Combine 2y^{2} and y^{2} to get 3y^{2}.

$$3y^{2}+60y+400=101$$

Subtract 101 from both sides.

$$3y^{2}+60y+400-101=0$$

Subtract 101 from 400 to get 299.

$$3y^{2}+60y+299=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 60 for b, and 299 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-60±\sqrt{60^{2}-4\times 3\times 299}}{2\times 3}$$

Square 60.

$$y=\frac{-60±\sqrt{3600-4\times 3\times 299}}{2\times 3}$$

Multiply -4 times 3.

$$y=\frac{-60±\sqrt{3600-12\times 299}}{2\times 3}$$

Multiply -12 times 299.

$$y=\frac{-60±\sqrt{3600-3588}}{2\times 3}$$

Add 3600 to -3588.

$$y=\frac{-60±\sqrt{12}}{2\times 3}$$

Take the square root of 12.

$$y=\frac{-60±2\sqrt{3}}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{-60±2\sqrt{3}}{6}$$

Now solve the equation y=\frac{-60±2\sqrt{3}}{6} when ± is plus. Add -60 to 2\sqrt{3}.

$$y=\frac{2\sqrt{3}-60}{6}$$

Divide -60+2\sqrt{3} by 6.

$$y=\frac{\sqrt{3}}{3}-10$$

Now solve the equation y=\frac{-60±2\sqrt{3}}{6} when ± is minus. Subtract 2\sqrt{3} from -60.

$$y=\frac{-2\sqrt{3}-60}{6}$$

Divide -60-2\sqrt{3} by 6.

$$y=-\frac{\sqrt{3}}{3}-10$$

The equation is now solved.

$$y=\frac{\sqrt{3}}{3}-10 y=-\frac{\sqrt{3}}{3}-10$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+20\right)^{2}.

$$y^{2}+40y+400+y^{2}+20y+y^{2}=101$$

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

$$2y^{2}+40y+400+20y+y^{2}=101$$

Combine 40y and 20y to get 60y.

$$2y^{2}+60y+400+y^{2}=101$$

Combine 2y^{2} and y^{2} to get 3y^{2}.

$$3y^{2}+60y+400=101$$

Subtract 400 from both sides.

$$3y^{2}+60y=101-400$$

Subtract 400 from 101 to get -299.

$$3y^{2}+60y=-299$$

Divide both sides by 3.

$$\frac{3y^{2}+60y}{3}=-\frac{299}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$y^{2}+\frac{60}{3}y=-\frac{299}{3}$$

Divide 60 by 3.

$$y^{2}+20y=-\frac{299}{3}$$

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.

$$y^{2}+20y+10^{2}=-\frac{299}{3}+10^{2}$$

Square 10.

$$y^{2}+20y+100=-\frac{299}{3}+100$$

Add -\frac{299}{3} to 100.

$$y^{2}+20y+100=\frac{1}{3}$$

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}.

$$\left(y+10\right)^{2}=\frac{1}{3}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+10\right)^{2}}=\sqrt{\frac{1}{3}}$$

Simplify.

$$y+10=\frac{\sqrt{3}}{3} y+10=-\frac{\sqrt{3}}{3}$$

Subtract 10 from both sides of the equation.

$$y=\frac{\sqrt{3}}{3}-10 y=-\frac{\sqrt{3}}{3}-10$$