Solve 2x+8x^2=2-2-7+9

$$2x+8 { x }^{ 2 } =2-2-7+9$$

Solve for x

$$x=\frac{\sqrt{17}-1}{8}\approx 0.390388203$$

$$x=\frac{-\sqrt{17}-1}{8}\approx -0.640388203$$

Steps Using the Quadratic Formula

Subtract 2 from 2 to get 0.

$$2x+8x^{2}=-7+9$$

Add -7 and 9 to get 2.

$$2x+8x^{2}=2$$

Subtract 2 from both sides.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 2 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 8\left(-2\right)}}{2\times 8}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 8\left(-2\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-2±\sqrt{4-32\left(-2\right)}}{2\times 8}$$

Multiply -32 times -2.

$$x=\frac{-2±\sqrt{4+64}}{2\times 8}$$

Add 4 to 64.

$$x=\frac{-2±\sqrt{68}}{2\times 8}$$

Take the square root of 68.

$$x=\frac{-2±2\sqrt{17}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-2±2\sqrt{17}}{16}$$

Now solve the equation x=\frac{-2±2\sqrt{17}}{16} when ± is plus. Add -2 to 2\sqrt{17}.

$$x=\frac{2\sqrt{17}-2}{16}$$

Divide -2+2\sqrt{17} by 16.

$$x=\frac{\sqrt{17}-1}{8}$$

Now solve the equation x=\frac{-2±2\sqrt{17}}{16} when ± is minus. Subtract 2\sqrt{17} from -2.

$$x=\frac{-2\sqrt{17}-2}{16}$$

Divide -2-2\sqrt{17} by 16.

$$x=\frac{-\sqrt{17}-1}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{17}-1}{8} x=\frac{-\sqrt{17}-1}{8}$$

Steps for Completing the Square

Subtract 2 from 2 to get 0.

$$2x+8x^{2}=-7+9$$

Add -7 and 9 to get 2.

$$2x+8x^{2}=2$$

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.

$$8x^{2}+2x=2$$

Divide both sides by 8.

$$\frac{8x^{2}+2x}{8}=\frac{2}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\frac{2}{8}x=\frac{2}{8}$$

Reduce the fraction \frac{2}{8} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{1}{4}x=\frac{2}{8}$$

Reduce the fraction \frac{2}{8} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{1}{4}x=\frac{1}{4}$$

Divide \frac{1}{4}, the coefficient of the x term, by 2 to get \frac{1}{8}. Then add the square of \frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{4}x+\left(\frac{1}{8}\right)^{2}=\frac{1}{4}+\left(\frac{1}{8}\right)^{2}$$

Square \frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{1}{4}+\frac{1}{64}$$

Add \frac{1}{4} to \frac{1}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{4}x+\frac{1}{64}=\frac{17}{64}$$

Factor x^{2}+\frac{1}{4}x+\frac{1}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{8}\right)^{2}=\frac{17}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{8}\right)^{2}}=\sqrt{\frac{17}{64}}$$

Simplify.

$$x+\frac{1}{8}=\frac{\sqrt{17}}{8} x+\frac{1}{8}=-\frac{\sqrt{17}}{8}$$

Subtract \frac{1}{8} from both sides of the equation.

$$x=\frac{\sqrt{17}-1}{8} x=\frac{-\sqrt{17}-1}{8}$$