Solve y^2=10y-20

$$y ^ { 2 } = 10 y - 20$$

Solve for y

$$y=\sqrt{5}+5\approx 7.236067977$$

$$y=5-\sqrt{5}\approx 2.763932023$$

Steps Using the Quadratic Formula

Subtract 10y from both sides.

$$y^{2}-10y=-20$$

Add 20 to both sides.

$$y^{2}-10y+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 20}}{2}$$

Square -10.

$$y=\frac{-\left(-10\right)±\sqrt{100-4\times 20}}{2}$$

Multiply -4 times 20.

$$y=\frac{-\left(-10\right)±\sqrt{100-80}}{2}$$

Add 100 to -80.

$$y=\frac{-\left(-10\right)±\sqrt{20}}{2}$$

Take the square root of 20.

$$y=\frac{-\left(-10\right)±2\sqrt{5}}{2}$$

The opposite of -10 is 10.

$$y=\frac{10±2\sqrt{5}}{2}$$

Now solve the equation y=\frac{10±2\sqrt{5}}{2} when ± is plus. Add 10 to 2\sqrt{5}.

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

Divide 10+2\sqrt{5} by 2.

$$y=\sqrt{5}+5$$

Now solve the equation y=\frac{10±2\sqrt{5}}{2} when ± is minus. Subtract 2\sqrt{5} from 10.

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

Divide 10-2\sqrt{5} by 2.

$$y=5-\sqrt{5}$$

The equation is now solved.

$$y=\sqrt{5}+5 y=5-\sqrt{5}$$

Steps for Completing the Square

Subtract 10y from both sides.

$$y^{2}-10y=-20$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-10y+\left(-5\right)^{2}=-20+\left(-5\right)^{2}$$

Square -5.

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

Add -20 to 25.

$$y^{2}-10y+25=5$$

Factor y^{2}-10y+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-5\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-5\right)^{2}}=\sqrt{5}$$

Simplify.

$$y-5=\sqrt{5} y-5=-\sqrt{5}$$

Add 5 to both sides of the equation.

$$y=\sqrt{5}+5 y=5-\sqrt{5}$$