Solve x^2-25x-4500=0

$${ x }^{ 2 } -25x-4500=0$$

Solve for x

$$x = \frac{5 \sqrt{745} + 25}{2} \approx 80.73672032$$

$$x=\frac{25-5\sqrt{745}}{2}\approx -55.73672032$$

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.

$$x^{2}-25x-4500=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -25 for b, and -4500 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\left(-4500\right)}}{2}$$

Square -25.

$$x=\frac{-\left(-25\right)±\sqrt{625-4\left(-4500\right)}}{2}$$

Multiply -4 times -4500.

$$x=\frac{-\left(-25\right)±\sqrt{625+18000}}{2}$$

Add 625 to 18000.

$$x=\frac{-\left(-25\right)±\sqrt{18625}}{2}$$

Take the square root of 18625.

$$x=\frac{-\left(-25\right)±5\sqrt{745}}{2}$$

The opposite of -25 is 25.

$$x=\frac{25±5\sqrt{745}}{2}$$

Now solve the equation x=\frac{25±5\sqrt{745}}{2} when ± is plus. Add 25 to 5\sqrt{745}.

$$x=\frac{5\sqrt{745}+25}{2}$$

Now solve the equation x=\frac{25±5\sqrt{745}}{2} when ± is minus. Subtract 5\sqrt{745} from 25.

$$x=\frac{25-5\sqrt{745}}{2}$$

The equation is now solved.

$$x=\frac{5\sqrt{745}+25}{2} x=\frac{25-5\sqrt{745}}{2}$$

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.

$$x^{2}-25x-4500=0$$

Add 4500 to both sides of the equation.

$$x^{2}-25x-4500-\left(-4500\right)=-\left(-4500\right)$$

Subtracting -4500 from itself leaves 0.

$$x^{2}-25x=-\left(-4500\right)$$

Subtract -4500 from 0.

$$x^{2}-25x=4500$$

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.

$$x^{2}-25x+\left(-\frac{25}{2}\right)^{2}=4500+\left(-\frac{25}{2}\right)^{2}$$

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

$$x^{2}-25x+\frac{625}{4}=4500+\frac{625}{4}$$

Add 4500 to \frac{625}{4}.

$$x^{2}-25x+\frac{625}{4}=\frac{18625}{4}$$

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}.

$$\left(x-\frac{25}{2}\right)^{2}=\frac{18625}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{25}{2}\right)^{2}}=\sqrt{\frac{18625}{4}}$$

Simplify.

$$x-\frac{25}{2}=\frac{5\sqrt{745}}{2} x-\frac{25}{2}=-\frac{5\sqrt{745}}{2}$$

Add \frac{25}{2} to both sides of the equation.

$$x=\frac{5\sqrt{745}+25}{2} x=\frac{25-5\sqrt{745}}{2}$$