Solve left(x+5right)^2=-20(y-95)

$${ \left(x+5 \right) }^{ 2 } =-20(y-95)$$

Solve for y

$$y=-\frac{\left(x+5\right)^{2}}{20}+95$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$x^{2}+10x+25=-20\left(y-95\right)$$

Use the distributive property to multiply -20 by y-95.

$$x^{2}+10x+25=-20y+1900$$

Swap sides so that all variable terms are on the left hand side.

$$-20y+1900=x^{2}+10x+25$$

Subtract 1900 from both sides.

$$-20y=x^{2}+10x+25-1900$$

Subtract 1900 from 25 to get -1875.

$$-20y=x^{2}+10x-1875$$

Divide both sides by -20.

$$\frac{-20y}{-20}=\frac{x^{2}+10x-1875}{-20}$$

Dividing by -20 undoes the multiplication by -20.

$$y=\frac{x^{2}+10x-1875}{-20}$$

Divide x^{2}+10x-1875 by -20.

$$y=-\frac{x^{2}}{20}-\frac{x}{2}+\frac{375}{4}$$

Solve for x (complex solution)

$$x=-2\sqrt{475-5y}-5$$

$$x=2\sqrt{475-5y}-5$$

Solve for x

$$x=-2\sqrt{475-5y}-5$$

$$x=2\sqrt{475-5y}-5\text{, }y\leq 95$$