Solve (60-35)*100(y+11)^2=160-35+11)*100

$$( 60 - 35 ) \cdot 100 ( y + 11 ) ^ { 2 } = 160 - 35 + 11 ) \times 100$$

Solve for y

$$y=\frac{\sqrt{34}}{25}-11\approx -10.766761924$$

$$y=-\frac{\sqrt{34}}{25}-11\approx -11.233238076$$

Steps Using the Quadratic Formula

Subtract 35 from 60 to get 25.

$$25\times 100\left(y+11\right)^{2}=160-35+11$$

Multiply 25 and 100 to get 2500.

$$2500\left(y+11\right)^{2}=160-35+11$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+11\right)^{2}.

$$2500\left(y^{2}+22y+121\right)=160-35+11$$

Use the distributive property to multiply 2500 by y^{2}+22y+121.

$$2500y^{2}+55000y+302500=160-35+11$$

Subtract 35 from 160 to get 125.

$$2500y^{2}+55000y+302500=125+11$$

Add 125 and 11 to get 136.

$$2500y^{2}+55000y+302500=136$$

Subtract 136 from both sides.

$$2500y^{2}+55000y+302500-136=0$$

Subtract 136 from 302500 to get 302364.

$$2500y^{2}+55000y+302364=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2500 for a, 55000 for b, and 302364 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-55000±\sqrt{55000^{2}-4\times 2500\times 302364}}{2\times 2500}$$

Square 55000.

$$y=\frac{-55000±\sqrt{3025000000-4\times 2500\times 302364}}{2\times 2500}$$

Multiply -4 times 2500.

$$y=\frac{-55000±\sqrt{3025000000-10000\times 302364}}{2\times 2500}$$

Multiply -10000 times 302364.

$$y=\frac{-55000±\sqrt{3025000000-3023640000}}{2\times 2500}$$

Add 3025000000 to -3023640000.

$$y=\frac{-55000±\sqrt{1360000}}{2\times 2500}$$

Take the square root of 1360000.

$$y=\frac{-55000±200\sqrt{34}}{2\times 2500}$$

Multiply 2 times 2500.

$$y=\frac{-55000±200\sqrt{34}}{5000}$$

Now solve the equation y=\frac{-55000±200\sqrt{34}}{5000} when ± is plus. Add -55000 to 200\sqrt{34}.

$$y=\frac{200\sqrt{34}-55000}{5000}$$

Divide -55000+200\sqrt{34} by 5000.

$$y=\frac{\sqrt{34}}{25}-11$$

Now solve the equation y=\frac{-55000±200\sqrt{34}}{5000} when ± is minus. Subtract 200\sqrt{34} from -55000.

$$y=\frac{-200\sqrt{34}-55000}{5000}$$

Divide -55000-200\sqrt{34} by 5000.

$$y=-\frac{\sqrt{34}}{25}-11$$

The equation is now solved.

$$y=\frac{\sqrt{34}}{25}-11 y=-\frac{\sqrt{34}}{25}-11$$

Steps for Completing the Square

Subtract 35 from 60 to get 25.

$$25\times 100\left(y+11\right)^{2}=160-35+11$$

Multiply 25 and 100 to get 2500.

$$2500\left(y+11\right)^{2}=160-35+11$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+11\right)^{2}.

$$2500\left(y^{2}+22y+121\right)=160-35+11$$

Use the distributive property to multiply 2500 by y^{2}+22y+121.

$$2500y^{2}+55000y+302500=160-35+11$$

Subtract 35 from 160 to get 125.

$$2500y^{2}+55000y+302500=125+11$$

Add 125 and 11 to get 136.

$$2500y^{2}+55000y+302500=136$$

Subtract 302500 from both sides.

$$2500y^{2}+55000y=136-302500$$

Subtract 302500 from 136 to get -302364.

$$2500y^{2}+55000y=-302364$$

Divide both sides by 2500.

$$\frac{2500y^{2}+55000y}{2500}=-\frac{302364}{2500}$$

Dividing by 2500 undoes the multiplication by 2500.

$$y^{2}+\frac{55000}{2500}y=-\frac{302364}{2500}$$

Divide 55000 by 2500.

$$y^{2}+22y=-\frac{302364}{2500}$$

Reduce the fraction \frac{-302364}{2500} to lowest terms by extracting and canceling out 4.

$$y^{2}+22y=-\frac{75591}{625}$$

Divide 22, the coefficient of the x term, by 2 to get 11. Then add the square of 11 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+22y+11^{2}=-\frac{75591}{625}+11^{2}$$

Square 11.

$$y^{2}+22y+121=-\frac{75591}{625}+121$$

Add -\frac{75591}{625} to 121.

$$y^{2}+22y+121=\frac{34}{625}$$

Factor y^{2}+22y+121. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+11\right)^{2}=\frac{34}{625}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+11\right)^{2}}=\sqrt{\frac{34}{625}}$$

Simplify.

$$y+11=\frac{\sqrt{34}}{25} y+11=-\frac{\sqrt{34}}{25}$$

Subtract 11 from both sides of the equation.

$$y=\frac{\sqrt{34}}{25}-11 y=-\frac{\sqrt{34}}{25}-11$$