Solve x-10x+2x^2=58

$$x-10x+2 { x }^{ 2 } =58$$

Solve for x

$$x = \frac{\sqrt{545} + 9}{4} \approx 8.086308765$$

$$x=\frac{9-\sqrt{545}}{4}\approx -3.586308765$$

Steps Using the Quadratic Formula

Combine x and -10x to get -9x.

$$-9x+2x^{2}=58$$

Subtract 58 from both sides.

$$-9x+2x^{2}-58=0$$

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.

$$2x^{2}-9x-58=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -9 for b, and -58 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 2\left(-58\right)}}{2\times 2}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 2\left(-58\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-9\right)±\sqrt{81-8\left(-58\right)}}{2\times 2}$$

Multiply -8 times -58.

$$x=\frac{-\left(-9\right)±\sqrt{81+464}}{2\times 2}$$

Add 81 to 464.

$$x=\frac{-\left(-9\right)±\sqrt{545}}{2\times 2}$$

The opposite of -9 is 9.

$$x=\frac{9±\sqrt{545}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{9±\sqrt{545}}{4}$$

Now solve the equation x=\frac{9±\sqrt{545}}{4} when ± is plus. Add 9 to \sqrt{545}.

$$x=\frac{\sqrt{545}+9}{4}$$

Now solve the equation x=\frac{9±\sqrt{545}}{4} when ± is minus. Subtract \sqrt{545} from 9.

$$x=\frac{9-\sqrt{545}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{545}+9}{4} x=\frac{9-\sqrt{545}}{4}$$

Steps for Completing the Square

Combine x and -10x to get -9x.

$$-9x+2x^{2}=58$$

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.

$$2x^{2}-9x=58$$

Divide both sides by 2.

$$\frac{2x^{2}-9x}{2}=\frac{58}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{9}{2}x=\frac{58}{2}$$

Divide 58 by 2.

$$x^{2}-\frac{9}{2}x=29$$

Divide -\frac{9}{2}, the coefficient of the x term, by 2 to get -\frac{9}{4}. Then add the square of -\frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{9}{2}x+\left(-\frac{9}{4}\right)^{2}=29+\left(-\frac{9}{4}\right)^{2}$$

Square -\frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=29+\frac{81}{16}$$

Add 29 to \frac{81}{16}.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=\frac{545}{16}$$

Factor x^{2}-\frac{9}{2}x+\frac{81}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{4}\right)^{2}=\frac{545}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{4}\right)^{2}}=\sqrt{\frac{545}{16}}$$

Simplify.

$$x-\frac{9}{4}=\frac{\sqrt{545}}{4} x-\frac{9}{4}=-\frac{\sqrt{545}}{4}$$

Add \frac{9}{4} to both sides of the equation.

$$x=\frac{\sqrt{545}+9}{4} x=\frac{9-\sqrt{545}}{4}$$