Solve 12x+2=8x^2

$$12 x + 2 = 8 x ^ { 2 }$$

Solve for x

$$x = \frac{\sqrt{13} + 3}{4} \approx 1.651387819$$

$$x=\frac{3-\sqrt{13}}{4}\approx -0.151387819$$

Steps Using the Quadratic Formula

Subtract 8x^{2} from both sides.

$$12x+2-8x^{2}=0$$

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.

$$-8x^{2}+12x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, 12 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-8\right)\times 2}}{2\left(-8\right)}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-8\right)\times 2}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$x=\frac{-12±\sqrt{144+32\times 2}}{2\left(-8\right)}$$

Multiply 32 times 2.

$$x=\frac{-12±\sqrt{144+64}}{2\left(-8\right)}$$

Add 144 to 64.

$$x=\frac{-12±\sqrt{208}}{2\left(-8\right)}$$

Take the square root of 208.

$$x=\frac{-12±4\sqrt{13}}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{-12±4\sqrt{13}}{-16}$$

Now solve the equation x=\frac{-12±4\sqrt{13}}{-16} when ± is plus. Add -12 to 4\sqrt{13}.

$$x=\frac{4\sqrt{13}-12}{-16}$$

Divide -12+4\sqrt{13} by -16.

$$x=\frac{3-\sqrt{13}}{4}$$

Now solve the equation x=\frac{-12±4\sqrt{13}}{-16} when ± is minus. Subtract 4\sqrt{13} from -12.

$$x=\frac{-4\sqrt{13}-12}{-16}$$

Divide -12-4\sqrt{13} by -16.

$$x=\frac{\sqrt{13}+3}{4}$$

The equation is now solved.

$$x=\frac{3-\sqrt{13}}{4} x=\frac{\sqrt{13}+3}{4}$$

Steps for Completing the Square

Subtract 8x^{2} from both sides.

$$12x+2-8x^{2}=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$12x-8x^{2}=-2$$

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.

$$-8x^{2}+12x=-2$$

Divide both sides by -8.

$$\frac{-8x^{2}+12x}{-8}=-\frac{2}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$x^{2}+\frac{12}{-8}x=-\frac{2}{-8}$$

Reduce the fraction \frac{12}{-8} to lowest terms by extracting and canceling out 4.

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

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

$$x^{2}-\frac{3}{2}x=\frac{1}{4}$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=\frac{1}{4}+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{1}{4}+\frac{9}{16}$$

Add \frac{1}{4} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{13}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{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}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{13}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{13}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{\sqrt{13}}{4} x-\frac{3}{4}=-\frac{\sqrt{13}}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{\sqrt{13}+3}{4} x=\frac{3-\sqrt{13}}{4}$$