Solve 5x^2-125=50x

$$5 { x }^{ 2 } -125=50x$$

Solve for x

$$x=5\sqrt{2}+5\approx 12.071067812$$

$$x=5-5\sqrt{2}\approx -2.071067812$$

Steps Using the Quadratic Formula

Subtract 50x from both sides.

$$5x^{2}-125-50x=0$$

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.

$$5x^{2}-50x-125=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -50 for b, and -125 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-50\right)±\sqrt{\left(-50\right)^{2}-4\times 5\left(-125\right)}}{2\times 5}$$

Square -50.

$$x=\frac{-\left(-50\right)±\sqrt{2500-4\times 5\left(-125\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-50\right)±\sqrt{2500-20\left(-125\right)}}{2\times 5}$$

Multiply -20 times -125.

$$x=\frac{-\left(-50\right)±\sqrt{2500+2500}}{2\times 5}$$

Add 2500 to 2500.

$$x=\frac{-\left(-50\right)±\sqrt{5000}}{2\times 5}$$

Take the square root of 5000.

$$x=\frac{-\left(-50\right)±50\sqrt{2}}{2\times 5}$$

The opposite of -50 is 50.

$$x=\frac{50±50\sqrt{2}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{50±50\sqrt{2}}{10}$$

Now solve the equation x=\frac{50±50\sqrt{2}}{10} when ± is plus. Add 50 to 50\sqrt{2}.

$$x=\frac{50\sqrt{2}+50}{10}$$

Divide 50+50\sqrt{2} by 10.

$$x=5\sqrt{2}+5$$

Now solve the equation x=\frac{50±50\sqrt{2}}{10} when ± is minus. Subtract 50\sqrt{2} from 50.

$$x=\frac{50-50\sqrt{2}}{10}$$

Divide 50-50\sqrt{2} by 10.

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

The equation is now solved.

$$x=5\sqrt{2}+5 x=5-5\sqrt{2}$$

Steps for Completing the Square

Subtract 50x from both sides.

$$5x^{2}-125-50x=0$$

Add 125 to both sides. Anything plus zero gives itself.

$$5x^{2}-50x=125$$

Divide both sides by 5.

$$\frac{5x^{2}-50x}{5}=\frac{125}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\left(-\frac{50}{5}\right)x=\frac{125}{5}$$

Divide -50 by 5.

$$x^{2}-10x=\frac{125}{5}$$

Divide 125 by 5.

$$x^{2}-10x=25$$

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.

$$x^{2}-10x+\left(-5\right)^{2}=25+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=25+25$$

Add 25 to 25.

$$x^{2}-10x+25=50$$

Factor x^{2}-10x+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}.

$$\left(x-5\right)^{2}=50$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{50}$$

Simplify.

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

Add 5 to both sides of the equation.

$$x=5\sqrt{2}+5 x=5-5\sqrt{2}$$