Solve 25x^2-x+10=x

$$25 x ^ { 2 } - x + 10 = x$$

Solve for x (complex solution)

$$x=\frac{1+\sqrt{249}i}{25}\approx 0.04+0.631189354i$$

$$x=\frac{-\sqrt{249}i+1}{25}\approx 0.04-0.631189354i$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$25x^{2}-x+10-x=0$$

Combine -x and -x to get -2x.

$$25x^{2}-2x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, -2 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 25\times 10}}{2\times 25}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 25\times 10}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-\left(-2\right)±\sqrt{4-100\times 10}}{2\times 25}$$

Multiply -100 times 10.

$$x=\frac{-\left(-2\right)±\sqrt{4-1000}}{2\times 25}$$

Add 4 to -1000.

$$x=\frac{-\left(-2\right)±\sqrt{-996}}{2\times 25}$$

Take the square root of -996.

$$x=\frac{-\left(-2\right)±2\sqrt{249}i}{2\times 25}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{249}i}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{2±2\sqrt{249}i}{50}$$

Now solve the equation x=\frac{2±2\sqrt{249}i}{50} when ± is plus. Add 2 to 2i\sqrt{249}.

$$x=\frac{2+2\sqrt{249}i}{50}$$

Divide 2+2i\sqrt{249} by 50.

$$x=\frac{1+\sqrt{249}i}{25}$$

Now solve the equation x=\frac{2±2\sqrt{249}i}{50} when ± is minus. Subtract 2i\sqrt{249} from 2.

$$x=\frac{-2\sqrt{249}i+2}{50}$$

Divide 2-2i\sqrt{249} by 50.

$$x=\frac{-\sqrt{249}i+1}{25}$$

The equation is now solved.

$$x=\frac{1+\sqrt{249}i}{25} x=\frac{-\sqrt{249}i+1}{25}$$

Steps for Completing the Square

Subtract x from both sides.

$$25x^{2}-x+10-x=0$$

Combine -x and -x to get -2x.

$$25x^{2}-2x+10=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$25x^{2}-2x=-10$$

Divide both sides by 25.

$$\frac{25x^{2}-2x}{25}=-\frac{10}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$x^{2}-\frac{2}{25}x=-\frac{10}{25}$$

Reduce the fraction \frac{-10}{25} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{2}{25}x=-\frac{2}{5}$$

Divide -\frac{2}{25}, the coefficient of the x term, by 2 to get -\frac{1}{25}. Then add the square of -\frac{1}{25} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{25} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{25}x+\frac{1}{625}=-\frac{2}{5}+\frac{1}{625}$$

Add -\frac{2}{5} to \frac{1}{625} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{25}x+\frac{1}{625}=-\frac{249}{625}$$

Factor x^{2}-\frac{2}{25}x+\frac{1}{625}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{25}\right)^{2}=-\frac{249}{625}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{25}\right)^{2}}=\sqrt{-\frac{249}{625}}$$

Simplify.

$$x-\frac{1}{25}=\frac{\sqrt{249}i}{25} x-\frac{1}{25}=-\frac{\sqrt{249}i}{25}$$

Add \frac{1}{25} to both sides of the equation.

$$x=\frac{1+\sqrt{249}i}{25} x=\frac{-\sqrt{249}i+1}{25}$$