Solve x^2+25x-125=0

$${ x }^{ 2 } +25x-125=0$$

Solve for x

$$x = \frac{15 \sqrt{5} - 25}{2} \approx 4.270509831$$

$$x=\frac{-15\sqrt{5}-25}{2}\approx -29.270509831$$

Steps Using the Quadratic Formula

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.

$$x^{2}+25x-125=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 25 for b, and -125 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-25±\sqrt{25^{2}-4\left(-125\right)}}{2}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\left(-125\right)}}{2}$$

Multiply -4 times -125.

$$x=\frac{-25±\sqrt{625+500}}{2}$$

Add 625 to 500.

$$x=\frac{-25±\sqrt{1125}}{2}$$

Take the square root of 1125.

$$x=\frac{-25±15\sqrt{5}}{2}$$

Now solve the equation x=\frac{-25±15\sqrt{5}}{2} when ± is plus. Add -25 to 15\sqrt{5}.

$$x=\frac{15\sqrt{5}-25}{2}$$

Now solve the equation x=\frac{-25±15\sqrt{5}}{2} when ± is minus. Subtract 15\sqrt{5} from -25.

$$x=\frac{-15\sqrt{5}-25}{2}$$

The equation is now solved.

$$x=\frac{15\sqrt{5}-25}{2} x=\frac{-15\sqrt{5}-25}{2}$$

Steps for Completing the Square

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.

$$x^{2}+25x-125=0$$

Add 125 to both sides of the equation.

$$x^{2}+25x-125-\left(-125\right)=-\left(-125\right)$$

Subtracting -125 from itself leaves 0.

$$x^{2}+25x=-\left(-125\right)$$

Subtract -125 from 0.

$$x^{2}+25x=125$$

Divide 25, the coefficient of the x term, by 2 to get \frac{25}{2}. Then add the square of \frac{25}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+25x+\left(\frac{25}{2}\right)^{2}=125+\left(\frac{25}{2}\right)^{2}$$

Square \frac{25}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+25x+\frac{625}{4}=125+\frac{625}{4}$$

Add 125 to \frac{625}{4}.

$$x^{2}+25x+\frac{625}{4}=\frac{1125}{4}$$

Factor x^{2}+25x+\frac{625}{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}.

$$\left(x+\frac{25}{2}\right)^{2}=\frac{1125}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{25}{2}\right)^{2}}=\sqrt{\frac{1125}{4}}$$

Simplify.

$$x+\frac{25}{2}=\frac{15\sqrt{5}}{2} x+\frac{25}{2}=-\frac{15\sqrt{5}}{2}$$

Subtract \frac{25}{2} from both sides of the equation.

$$x=\frac{15\sqrt{5}-25}{2} x=\frac{-15\sqrt{5}-25}{2}$$