Solve y^2=10y+3

$$y ^ { 2 } = 10 y + 3$$

Solve for y

$$y=2\sqrt{7}+5\approx 10.291502622$$

$$y=5-2\sqrt{7}\approx -0.291502622$$

Steps Using the Quadratic Formula

Subtract 10y from both sides.

$$y^{2}-10y=3$$

Subtract 3 from both sides.

$$y^{2}-10y-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-3\right)}}{2}$$

Square -10.

$$y=\frac{-\left(-10\right)±\sqrt{100-4\left(-3\right)}}{2}$$

Multiply -4 times -3.

$$y=\frac{-\left(-10\right)±\sqrt{100+12}}{2}$$

Add 100 to 12.

$$y=\frac{-\left(-10\right)±\sqrt{112}}{2}$$

Take the square root of 112.

$$y=\frac{-\left(-10\right)±4\sqrt{7}}{2}$$

The opposite of -10 is 10.

$$y=\frac{10±4\sqrt{7}}{2}$$

Now solve the equation y=\frac{10±4\sqrt{7}}{2} when ± is plus. Add 10 to 4\sqrt{7}.

$$y=\frac{4\sqrt{7}+10}{2}$$

Divide 10+4\sqrt{7} by 2.

$$y=2\sqrt{7}+5$$

Now solve the equation y=\frac{10±4\sqrt{7}}{2} when ± is minus. Subtract 4\sqrt{7} from 10.

$$y=\frac{10-4\sqrt{7}}{2}$$

Divide 10-4\sqrt{7} by 2.

$$y=5-2\sqrt{7}$$

The equation is now solved.

$$y=2\sqrt{7}+5 y=5-2\sqrt{7}$$

Steps for Completing the Square

Subtract 10y from both sides.

$$y^{2}-10y=3$$

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.

$$y^{2}-10y+\left(-5\right)^{2}=3+\left(-5\right)^{2}$$

Square -5.

$$y^{2}-10y+25=3+25$$

Add 3 to 25.

$$y^{2}-10y+25=28$$

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}.

$$\left(y-5\right)^{2}=28$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-5\right)^{2}}=\sqrt{28}$$

Simplify.

$$y-5=2\sqrt{7} y-5=-2\sqrt{7}$$

Add 5 to both sides of the equation.

$$y=2\sqrt{7}+5 y=5-2\sqrt{7}$$