Solve 3x+6=8x^2-2

$$3 x + 6 = 8 x ^ { 2 } - 2$$

Solve for x

$$x = \frac{\sqrt{265} + 3}{16} \approx 1.204926287$$

$$x=\frac{3-\sqrt{265}}{16}\approx -0.829926287$$

Steps Using the Quadratic Formula

Subtract 8x^{2} from both sides.

$$3x+6-8x^{2}=-2$$

Add 2 to both sides.

$$3x+6-8x^{2}+2=0$$

Add 6 and 2 to get 8.

$$3x+8-8x^{2}=0$$

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.

$$-8x^{2}+3x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, 3 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-8\right)\times 8}}{2\left(-8\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-8\right)\times 8}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$x=\frac{-3±\sqrt{9+32\times 8}}{2\left(-8\right)}$$

Multiply 32 times 8.

$$x=\frac{-3±\sqrt{9+256}}{2\left(-8\right)}$$

Add 9 to 256.

$$x=\frac{-3±\sqrt{265}}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{-3±\sqrt{265}}{-16}$$

Now solve the equation x=\frac{-3±\sqrt{265}}{-16} when ± is plus. Add -3 to \sqrt{265}.

$$x=\frac{\sqrt{265}-3}{-16}$$

Divide -3+\sqrt{265} by -16.

$$x=\frac{3-\sqrt{265}}{16}$$

Now solve the equation x=\frac{-3±\sqrt{265}}{-16} when ± is minus. Subtract \sqrt{265} from -3.

$$x=\frac{-\sqrt{265}-3}{-16}$$

Divide -3-\sqrt{265} by -16.

$$x=\frac{\sqrt{265}+3}{16}$$

The equation is now solved.

$$x=\frac{3-\sqrt{265}}{16} x=\frac{\sqrt{265}+3}{16}$$

Steps for Completing the Square

Subtract 8x^{2} from both sides.

$$3x+6-8x^{2}=-2$$

Subtract 6 from both sides.

$$3x-8x^{2}=-2-6$$

Subtract 6 from -2 to get -8.

$$3x-8x^{2}=-8$$

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.

$$-8x^{2}+3x=-8$$

Divide both sides by -8.

$$\frac{-8x^{2}+3x}{-8}=-\frac{8}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$x^{2}+\frac{3}{-8}x=-\frac{8}{-8}$$

Divide 3 by -8.

$$x^{2}-\frac{3}{8}x=-\frac{8}{-8}$$

Divide -8 by -8.

$$x^{2}-\frac{3}{8}x=1$$

Divide -\frac{3}{8}, the coefficient of the x term, by 2 to get -\frac{3}{16}. Then add the square of -\frac{3}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{8}x+\left(-\frac{3}{16}\right)^{2}=1+\left(-\frac{3}{16}\right)^{2}$$

Square -\frac{3}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{8}x+\frac{9}{256}=1+\frac{9}{256}$$

Add 1 to \frac{9}{256}.

$$x^{2}-\frac{3}{8}x+\frac{9}{256}=\frac{265}{256}$$

Factor x^{2}-\frac{3}{8}x+\frac{9}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{16}\right)^{2}=\frac{265}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{16}\right)^{2}}=\sqrt{\frac{265}{256}}$$

Simplify.

$$x-\frac{3}{16}=\frac{\sqrt{265}}{16} x-\frac{3}{16}=-\frac{\sqrt{265}}{16}$$

Add \frac{3}{16} to both sides of the equation.

$$x=\frac{\sqrt{265}+3}{16} x=\frac{3-\sqrt{265}}{16}$$