Solve -(8x+3)=3x(4x)-6

$$- ( 8 x + 3 ) = 3 x ( 4 x ) - 6$$

Solve for x

$$x=\frac{\sqrt{13}}{6}-\frac{1}{3}\approx 0.267591879$$

$$x=-\frac{\sqrt{13}}{6}-\frac{1}{3}\approx -0.934258546$$

Steps Using the Quadratic Formula

Multiply x and x to get x^{2}.

$$-\left(8x+3\right)=3x^{2}\times 4-6$$

To find the opposite of 8x+3, find the opposite of each term.

$$-8x-3=3x^{2}\times 4-6$$

Multiply 3 and 4 to get 12.

$$-8x-3=12x^{2}-6$$

Subtract 12x^{2} from both sides.

$$-8x-3-12x^{2}=-6$$

Add 6 to both sides.

$$-8x-3-12x^{2}+6=0$$

Add -3 and 6 to get 3.

$$-8x+3-12x^{2}=0$$

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.

$$-12x^{2}-8x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -12 for a, -8 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-12\right)\times 3}}{2\left(-12\right)}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-12\right)\times 3}}{2\left(-12\right)}$$

Multiply -4 times -12.

$$x=\frac{-\left(-8\right)±\sqrt{64+48\times 3}}{2\left(-12\right)}$$

Multiply 48 times 3.

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

Add 64 to 144.

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

Take the square root of 208.

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

The opposite of -8 is 8.

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

Multiply 2 times -12.

$$x=\frac{8±4\sqrt{13}}{-24}$$

Now solve the equation x=\frac{8±4\sqrt{13}}{-24} when ± is plus. Add 8 to 4\sqrt{13}.

$$x=\frac{4\sqrt{13}+8}{-24}$$

Divide 8+4\sqrt{13} by -24.

$$x=-\frac{\sqrt{13}}{6}-\frac{1}{3}$$

Now solve the equation x=\frac{8±4\sqrt{13}}{-24} when ± is minus. Subtract 4\sqrt{13} from 8.

$$x=\frac{8-4\sqrt{13}}{-24}$$

Divide 8-4\sqrt{13} by -24.

$$x=\frac{\sqrt{13}}{6}-\frac{1}{3}$$

The equation is now solved.

$$x=-\frac{\sqrt{13}}{6}-\frac{1}{3} x=\frac{\sqrt{13}}{6}-\frac{1}{3}$$

Steps for Completing the Square

Multiply x and x to get x^{2}.

$$-\left(8x+3\right)=3x^{2}\times 4-6$$

To find the opposite of 8x+3, find the opposite of each term.

$$-8x-3=3x^{2}\times 4-6$$

Multiply 3 and 4 to get 12.

$$-8x-3=12x^{2}-6$$

Subtract 12x^{2} from both sides.

$$-8x-3-12x^{2}=-6$$

Add 3 to both sides.

$$-8x-12x^{2}=-6+3$$

Add -6 and 3 to get -3.

$$-8x-12x^{2}=-3$$

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.

$$-12x^{2}-8x=-3$$

Divide both sides by -12.

$$\frac{-12x^{2}-8x}{-12}=-\frac{3}{-12}$$

Dividing by -12 undoes the multiplication by -12.

$$x^{2}+\left(-\frac{8}{-12}\right)x=-\frac{3}{-12}$$

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

$$x^{2}+\frac{2}{3}x=-\frac{3}{-12}$$

Reduce the fraction \frac{-3}{-12} to lowest terms by extracting and canceling out 3.

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

Divide \frac{2}{3}, the coefficient of the x term, by 2 to get \frac{1}{3}. Then add the square of \frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{3} by squaring both the numerator and the denominator of the fraction.

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

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

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{13}{36}$$

Factor x^{2}+\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{13}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{13}{36}}$$

Simplify.

$$x+\frac{1}{3}=\frac{\sqrt{13}}{6} x+\frac{1}{3}=-\frac{\sqrt{13}}{6}$$

Subtract \frac{1}{3} from both sides of the equation.

$$x=\frac{\sqrt{13}}{6}-\frac{1}{3} x=-\frac{\sqrt{13}}{6}-\frac{1}{3}$$