Solve x^2-45x-225=0

$${ x }^{ 2 } -45x-225=0$$

Solve for x

$$x = \frac{15 \sqrt{13} + 45}{2} \approx 49.541634566$$

$$x=\frac{45-15\sqrt{13}}{2}\approx -4.541634566$$

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.

$$x^{2}-45x-225=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -45 for b, and -225 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-45\right)±\sqrt{\left(-45\right)^{2}-4\left(-225\right)}}{2}$$

Square -45.

$$x=\frac{-\left(-45\right)±\sqrt{2025-4\left(-225\right)}}{2}$$

Multiply -4 times -225.

$$x=\frac{-\left(-45\right)±\sqrt{2025+900}}{2}$$

Add 2025 to 900.

$$x=\frac{-\left(-45\right)±\sqrt{2925}}{2}$$

Take the square root of 2925.

$$x=\frac{-\left(-45\right)±15\sqrt{13}}{2}$$

The opposite of -45 is 45.

$$x=\frac{45±15\sqrt{13}}{2}$$

Now solve the equation x=\frac{45±15\sqrt{13}}{2} when ± is plus. Add 45 to 15\sqrt{13}.

$$x=\frac{15\sqrt{13}+45}{2}$$

Now solve the equation x=\frac{45±15\sqrt{13}}{2} when ± is minus. Subtract 15\sqrt{13} from 45.

$$x=\frac{45-15\sqrt{13}}{2}$$

The equation is now solved.

$$x=\frac{15\sqrt{13}+45}{2} x=\frac{45-15\sqrt{13}}{2}$$

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.

$$x^{2}-45x-225=0$$

Add 225 to both sides of the equation.

$$x^{2}-45x-225-\left(-225\right)=-\left(-225\right)$$

Subtracting -225 from itself leaves 0.

$$x^{2}-45x=-\left(-225\right)$$

Subtract -225 from 0.

$$x^{2}-45x=225$$

Divide -45, the coefficient of the x term, by 2 to get -\frac{45}{2}. Then add the square of -\frac{45}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-45x+\left(-\frac{45}{2}\right)^{2}=225+\left(-\frac{45}{2}\right)^{2}$$

Square -\frac{45}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-45x+\frac{2025}{4}=225+\frac{2025}{4}$$

Add 225 to \frac{2025}{4}.

$$x^{2}-45x+\frac{2025}{4}=\frac{2925}{4}$$

Factor x^{2}-45x+\frac{2025}{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}.

$$\left(x-\frac{45}{2}\right)^{2}=\frac{2925}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{45}{2}\right)^{2}}=\sqrt{\frac{2925}{4}}$$

Simplify.

$$x-\frac{45}{2}=\frac{15\sqrt{13}}{2} x-\frac{45}{2}=-\frac{15\sqrt{13}}{2}$$

Add \frac{45}{2} to both sides of the equation.

$$x=\frac{15\sqrt{13}+45}{2} x=\frac{45-15\sqrt{13}}{2}$$