Solve (20-y)^2+y^2=25

$$( 20 - y ) ^ { 2 } + y ^ { 2 } = 25$$

Solve for y

$$y=\frac{5\sqrt{14}i}{2}+10\approx 10+9.354143467i$$

$$y=-\frac{5\sqrt{14}i}{2}+10\approx 10-9.354143467i$$

Steps Using the Quadratic Formula

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

$$400-40y+y^{2}+y^{2}=25$$

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

$$400-40y+2y^{2}=25$$

Subtract 25 from both sides.

$$400-40y+2y^{2}-25=0$$

Subtract 25 from 400 to get 375.

$$375-40y+2y^{2}=0$$

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.

$$2y^{2}-40y+375=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -40 for b, and 375 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 2\times 375}}{2\times 2}$$

Square -40.

$$y=\frac{-\left(-40\right)±\sqrt{1600-4\times 2\times 375}}{2\times 2}$$

Multiply -4 times 2.

$$y=\frac{-\left(-40\right)±\sqrt{1600-8\times 375}}{2\times 2}$$

Multiply -8 times 375.

$$y=\frac{-\left(-40\right)±\sqrt{1600-3000}}{2\times 2}$$

Add 1600 to -3000.

$$y=\frac{-\left(-40\right)±\sqrt{-1400}}{2\times 2}$$

Take the square root of -1400.

$$y=\frac{-\left(-40\right)±10\sqrt{14}i}{2\times 2}$$

The opposite of -40 is 40.

$$y=\frac{40±10\sqrt{14}i}{2\times 2}$$

Multiply 2 times 2.

$$y=\frac{40±10\sqrt{14}i}{4}$$

Now solve the equation y=\frac{40±10\sqrt{14}i}{4} when ± is plus. Add 40 to 10i\sqrt{14}.

$$y=\frac{40+10\sqrt{14}i}{4}$$

Divide 40+10i\sqrt{14} by 4.

$$y=\frac{5\sqrt{14}i}{2}+10$$

Now solve the equation y=\frac{40±10\sqrt{14}i}{4} when ± is minus. Subtract 10i\sqrt{14} from 40.

$$y=\frac{-10\sqrt{14}i+40}{4}$$

Divide 40-10i\sqrt{14} by 4.

$$y=-\frac{5\sqrt{14}i}{2}+10$$

The equation is now solved.

$$y=\frac{5\sqrt{14}i}{2}+10 y=-\frac{5\sqrt{14}i}{2}+10$$

Steps for Completing the Square

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

$$400-40y+y^{2}+y^{2}=25$$

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

$$400-40y+2y^{2}=25$$

Subtract 400 from both sides.

$$-40y+2y^{2}=25-400$$

Subtract 400 from 25 to get -375.

$$-40y+2y^{2}=-375$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2y^{2}-40y=-375$$

Divide both sides by 2.

$$\frac{2y^{2}-40y}{2}=-\frac{375}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$y^{2}+\left(-\frac{40}{2}\right)y=-\frac{375}{2}$$

Divide -40 by 2.

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

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.

$$y^{2}-20y+\left(-10\right)^{2}=-\frac{375}{2}+\left(-10\right)^{2}$$

Square -10.

$$y^{2}-20y+100=-\frac{375}{2}+100$$

Add -\frac{375}{2} to 100.

$$y^{2}-20y+100=-\frac{175}{2}$$

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}.

$$\left(y-10\right)^{2}=-\frac{175}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-10\right)^{2}}=\sqrt{-\frac{175}{2}}$$

Simplify.

$$y-10=\frac{5\sqrt{14}i}{2} y-10=-\frac{5\sqrt{14}i}{2}$$

Add 10 to both sides of the equation.

$$y=\frac{5\sqrt{14}i}{2}+10 y=-\frac{5\sqrt{14}i}{2}+10$$