Solve 2563+5*82=2x+5x^2

$$2563+5 \times 82=2x+5 { x }^{ 2 }$$

Solve for x

$$x = \frac{\sqrt{14866} - 1}{5} \approx 24.185241438$$

$$x=\frac{-\sqrt{14866}-1}{5}\approx -24.585241438$$

Steps Using the Quadratic Formula

Multiply 5 and 82 to get 410.

$$2563+410=2x+5x^{2}$$

Add 2563 and 410 to get 2973.

$$2973=2x+5x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$2x+5x^{2}=2973$$

Subtract 2973 from both sides.

$$2x+5x^{2}-2973=0$$

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.

$$5x^{2}+2x-2973=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, 2 for b, and -2973 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 5\left(-2973\right)}}{2\times 5}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 5\left(-2973\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-2±\sqrt{4-20\left(-2973\right)}}{2\times 5}$$

Multiply -20 times -2973.

$$x=\frac{-2±\sqrt{4+59460}}{2\times 5}$$

Add 4 to 59460.

$$x=\frac{-2±\sqrt{59464}}{2\times 5}$$

Take the square root of 59464.

$$x=\frac{-2±2\sqrt{14866}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-2±2\sqrt{14866}}{10}$$

Now solve the equation x=\frac{-2±2\sqrt{14866}}{10} when ± is plus. Add -2 to 2\sqrt{14866}.

$$x=\frac{2\sqrt{14866}-2}{10}$$

Divide -2+2\sqrt{14866} by 10.

$$x=\frac{\sqrt{14866}-1}{5}$$

Now solve the equation x=\frac{-2±2\sqrt{14866}}{10} when ± is minus. Subtract 2\sqrt{14866} from -2.

$$x=\frac{-2\sqrt{14866}-2}{10}$$

Divide -2-2\sqrt{14866} by 10.

$$x=\frac{-\sqrt{14866}-1}{5}$$

The equation is now solved.

$$x=\frac{\sqrt{14866}-1}{5} x=\frac{-\sqrt{14866}-1}{5}$$

Steps for Completing the Square

Multiply 5 and 82 to get 410.

$$2563+410=2x+5x^{2}$$

Add 2563 and 410 to get 2973.

$$2973=2x+5x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$2x+5x^{2}=2973$$

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.

$$5x^{2}+2x=2973$$

Divide both sides by 5.

$$\frac{5x^{2}+2x}{5}=\frac{2973}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}+\frac{2}{5}x=\frac{2973}{5}$$

Divide \frac{2}{5}, the coefficient of the x term, by 2 to get \frac{1}{5}. Then add the square of \frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{5}x+\left(\frac{1}{5}\right)^{2}=\frac{2973}{5}+\left(\frac{1}{5}\right)^{2}$$

Square \frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{5}x+\frac{1}{25}=\frac{2973}{5}+\frac{1}{25}$$

Add \frac{2973}{5} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{5}x+\frac{1}{25}=\frac{14866}{25}$$

Factor x^{2}+\frac{2}{5}x+\frac{1}{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}.

$$\left(x+\frac{1}{5}\right)^{2}=\frac{14866}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{5}\right)^{2}}=\sqrt{\frac{14866}{25}}$$

Simplify.

$$x+\frac{1}{5}=\frac{\sqrt{14866}}{5} x+\frac{1}{5}=-\frac{\sqrt{14866}}{5}$$

Subtract \frac{1}{5} from both sides of the equation.

$$x=\frac{\sqrt{14866}-1}{5} x=\frac{-\sqrt{14866}-1}{5}$$