Solve x^2-8x+10=13x

$$x ^ { 2 } - 8 x + 10 = 13 x$$

Solve for x

$$x = \frac{\sqrt{401} + 21}{2} \approx 20.512492197$$

$$x=\frac{21-\sqrt{401}}{2}\approx 0.487507803$$

Steps Using the Quadratic Formula

Subtract 13x from both sides.

$$x^{2}-8x+10-13x=0$$

Combine -8x and -13x to get -21x.

$$x^{2}-21x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -21 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-21\right)±\sqrt{\left(-21\right)^{2}-4\times 10}}{2}$$

Square -21.

$$x=\frac{-\left(-21\right)±\sqrt{441-4\times 10}}{2}$$

Multiply -4 times 10.

$$x=\frac{-\left(-21\right)±\sqrt{441-40}}{2}$$

Add 441 to -40.

$$x=\frac{-\left(-21\right)±\sqrt{401}}{2}$$

The opposite of -21 is 21.

$$x=\frac{21±\sqrt{401}}{2}$$

Now solve the equation x=\frac{21±\sqrt{401}}{2} when ± is plus. Add 21 to \sqrt{401}.

$$x=\frac{\sqrt{401}+21}{2}$$

Now solve the equation x=\frac{21±\sqrt{401}}{2} when ± is minus. Subtract \sqrt{401} from 21.

$$x=\frac{21-\sqrt{401}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{401}+21}{2} x=\frac{21-\sqrt{401}}{2}$$

Steps for Completing the Square

Subtract 13x from both sides.

$$x^{2}-8x+10-13x=0$$

Combine -8x and -13x to get -21x.

$$x^{2}-21x+10=0$$

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

$$x^{2}-21x=-10$$

Divide -21, the coefficient of the x term, by 2 to get -\frac{21}{2}. Then add the square of -\frac{21}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-21x+\left(-\frac{21}{2}\right)^{2}=-10+\left(-\frac{21}{2}\right)^{2}$$

Square -\frac{21}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-21x+\frac{441}{4}=-10+\frac{441}{4}$$

Add -10 to \frac{441}{4}.

$$x^{2}-21x+\frac{441}{4}=\frac{401}{4}$$

Factor x^{2}-21x+\frac{441}{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}.

$$\left(x-\frac{21}{2}\right)^{2}=\frac{401}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{21}{2}\right)^{2}}=\sqrt{\frac{401}{4}}$$

Simplify.

$$x-\frac{21}{2}=\frac{\sqrt{401}}{2} x-\frac{21}{2}=-\frac{\sqrt{401}}{2}$$

Add \frac{21}{2} to both sides of the equation.

$$x=\frac{\sqrt{401}+21}{2} x=\frac{21-\sqrt{401}}{2}$$