Solve 5y+10=yy-10

$$5 y + 10 = y y - 10$$

Solve for y

$$y = \frac{\sqrt{105} + 5}{2} \approx 7.623475383$$

$$y=\frac{5-\sqrt{105}}{2}\approx -2.623475383$$

Steps Using the Quadratic Formula

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

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

Subtract y^{2} from both sides.

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

Add 10 to both sides.

$$5y+10-y^{2}+10=0$$

Add 10 and 10 to get 20.

$$5y+20-y^{2}=0$$

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.

$$-y^{2}+5y+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 5 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\times 20}}{2\left(-1\right)}$$

Square 5.

$$y=\frac{-5±\sqrt{25-4\left(-1\right)\times 20}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$y=\frac{-5±\sqrt{25+4\times 20}}{2\left(-1\right)}$$

Multiply 4 times 20.

$$y=\frac{-5±\sqrt{25+80}}{2\left(-1\right)}$$

Add 25 to 80.

$$y=\frac{-5±\sqrt{105}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$y=\frac{-5±\sqrt{105}}{-2}$$

Now solve the equation y=\frac{-5±\sqrt{105}}{-2} when ± is plus. Add -5 to \sqrt{105}.

$$y=\frac{\sqrt{105}-5}{-2}$$

Divide -5+\sqrt{105} by -2.

$$y=\frac{5-\sqrt{105}}{2}$$

Now solve the equation y=\frac{-5±\sqrt{105}}{-2} when ± is minus. Subtract \sqrt{105} from -5.

$$y=\frac{-\sqrt{105}-5}{-2}$$

Divide -5-\sqrt{105} by -2.

$$y=\frac{\sqrt{105}+5}{2}$$

The equation is now solved.

$$y=\frac{5-\sqrt{105}}{2} y=\frac{\sqrt{105}+5}{2}$$

Steps for Completing the Square

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

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

Subtract y^{2} from both sides.

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

Subtract 10 from both sides.

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

Subtract 10 from -10 to get -20.

$$5y-y^{2}=-20$$

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.

$$-y^{2}+5y=-20$$

Divide both sides by -1.

$$\frac{-y^{2}+5y}{-1}=-\frac{20}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$y^{2}+\frac{5}{-1}y=-\frac{20}{-1}$$

Divide 5 by -1.

$$y^{2}-5y=-\frac{20}{-1}$$

Divide -20 by -1.

$$y^{2}-5y=20$$

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

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

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-5y+\frac{25}{4}=20+\frac{25}{4}$$

Add 20 to \frac{25}{4}.

$$y^{2}-5y+\frac{25}{4}=\frac{105}{4}$$

Factor y^{2}-5y+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{5}{2}\right)^{2}=\frac{105}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{5}{2}\right)^{2}}=\sqrt{\frac{105}{4}}$$

Simplify.

$$y-\frac{5}{2}=\frac{\sqrt{105}}{2} y-\frac{5}{2}=-\frac{\sqrt{105}}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$y=\frac{\sqrt{105}+5}{2} y=\frac{5-\sqrt{105}}{2}$$