Solve x^2-20=55x

$${ x }^{ 2 } -20 = 55x$$

Solve for x

$$x = \frac{3 \sqrt{345} + 55}{2} \approx 55.361263432$$

$$x=\frac{55-3\sqrt{345}}{2}\approx -0.361263432$$

Steps Using the Quadratic Formula

Subtract 55x from both sides.

$$x^{2}-20-55x=0$$

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.

$$x^{2}-55x-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -55 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-55\right)±\sqrt{\left(-55\right)^{2}-4\left(-20\right)}}{2}$$

Square -55.

$$x=\frac{-\left(-55\right)±\sqrt{3025-4\left(-20\right)}}{2}$$

Multiply -4 times -20.

$$x=\frac{-\left(-55\right)±\sqrt{3025+80}}{2}$$

Add 3025 to 80.

$$x=\frac{-\left(-55\right)±\sqrt{3105}}{2}$$

Take the square root of 3105.

$$x=\frac{-\left(-55\right)±3\sqrt{345}}{2}$$

The opposite of -55 is 55.

$$x=\frac{55±3\sqrt{345}}{2}$$

Now solve the equation x=\frac{55±3\sqrt{345}}{2} when ± is plus. Add 55 to 3\sqrt{345}.

$$x=\frac{3\sqrt{345}+55}{2}$$

Now solve the equation x=\frac{55±3\sqrt{345}}{2} when ± is minus. Subtract 3\sqrt{345} from 55.

$$x=\frac{55-3\sqrt{345}}{2}$$

The equation is now solved.

$$x=\frac{3\sqrt{345}+55}{2} x=\frac{55-3\sqrt{345}}{2}$$

Steps for Completing the Square

Subtract 55x from both sides.

$$x^{2}-20-55x=0$$

Add 20 to both sides. Anything plus zero gives itself.

$$x^{2}-55x=20$$

Divide -55, the coefficient of the x term, by 2 to get -\frac{55}{2}. Then add the square of -\frac{55}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-55x+\left(-\frac{55}{2}\right)^{2}=20+\left(-\frac{55}{2}\right)^{2}$$

Square -\frac{55}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-55x+\frac{3025}{4}=20+\frac{3025}{4}$$

Add 20 to \frac{3025}{4}.

$$x^{2}-55x+\frac{3025}{4}=\frac{3105}{4}$$

Factor x^{2}-55x+\frac{3025}{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}.

$$\left(x-\frac{55}{2}\right)^{2}=\frac{3105}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{55}{2}\right)^{2}}=\sqrt{\frac{3105}{4}}$$

Simplify.

$$x-\frac{55}{2}=\frac{3\sqrt{345}}{2} x-\frac{55}{2}=-\frac{3\sqrt{345}}{2}$$

Add \frac{55}{2} to both sides of the equation.

$$x=\frac{3\sqrt{345}+55}{2} x=\frac{55-3\sqrt{345}}{2}$$