Solve 3x-2(9(x))=6x^2+3x+4

$$3 x - 2 ( 9 ( x ) ) = 6 x ^ { 2 } + 3 x + 4$$

Solve for x

$$x=\frac{\sqrt{57}}{6}-\frac{3}{2}\approx -0.241694261$$

$$x=-\frac{\sqrt{57}}{6}-\frac{3}{2}\approx -2.758305739$$

Steps Using the Quadratic Formula

Multiply 2 and 9 to get 18.

$$3x-18x=6x^{2}+3x+4$$

Combine 3x and -18x to get -15x.

$$-15x=6x^{2}+3x+4$$

Subtract 6x^{2} from both sides.

$$-15x-6x^{2}=3x+4$$

Subtract 3x from both sides.

$$-15x-6x^{2}-3x=4$$

Combine -15x and -3x to get -18x.

$$-18x-6x^{2}=4$$

Subtract 4 from both sides.

$$-18x-6x^{2}-4=0$$

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.

$$-6x^{2}-18x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, -18 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\left(-6\right)\left(-4\right)}}{2\left(-6\right)}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\left(-6\right)\left(-4\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-\left(-18\right)±\sqrt{324+24\left(-4\right)}}{2\left(-6\right)}$$

Multiply 24 times -4.

$$x=\frac{-\left(-18\right)±\sqrt{324-96}}{2\left(-6\right)}$$

Add 324 to -96.

$$x=\frac{-\left(-18\right)±\sqrt{228}}{2\left(-6\right)}$$

Take the square root of 228.

$$x=\frac{-\left(-18\right)±2\sqrt{57}}{2\left(-6\right)}$$

The opposite of -18 is 18.

$$x=\frac{18±2\sqrt{57}}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{18±2\sqrt{57}}{-12}$$

Now solve the equation x=\frac{18±2\sqrt{57}}{-12} when ± is plus. Add 18 to 2\sqrt{57}.

$$x=\frac{2\sqrt{57}+18}{-12}$$

Divide 18+2\sqrt{57} by -12.

$$x=-\frac{\sqrt{57}}{6}-\frac{3}{2}$$

Now solve the equation x=\frac{18±2\sqrt{57}}{-12} when ± is minus. Subtract 2\sqrt{57} from 18.

$$x=\frac{18-2\sqrt{57}}{-12}$$

Divide 18-2\sqrt{57} by -12.

$$x=\frac{\sqrt{57}}{6}-\frac{3}{2}$$

The equation is now solved.

$$x=-\frac{\sqrt{57}}{6}-\frac{3}{2} x=\frac{\sqrt{57}}{6}-\frac{3}{2}$$

Steps for Completing the Square

Multiply 2 and 9 to get 18.

$$3x-18x=6x^{2}+3x+4$$

Combine 3x and -18x to get -15x.

$$-15x=6x^{2}+3x+4$$

Subtract 6x^{2} from both sides.

$$-15x-6x^{2}=3x+4$$

Subtract 3x from both sides.

$$-15x-6x^{2}-3x=4$$

Combine -15x and -3x to get -18x.

$$-18x-6x^{2}=4$$

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.

$$-6x^{2}-18x=4$$

Divide both sides by -6.

$$\frac{-6x^{2}-18x}{-6}=\frac{4}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\left(-\frac{18}{-6}\right)x=\frac{4}{-6}$$

Divide -18 by -6.

$$x^{2}+3x=\frac{4}{-6}$$

Reduce the fraction \frac{4}{-6} to lowest terms by extracting and canceling out 2.

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

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

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

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

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

Add -\frac{2}{3} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+3x+\frac{9}{4}=\frac{19}{12}$$

Factor x^{2}+3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{19}{12}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{19}{12}}$$

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{57}}{6} x+\frac{3}{2}=-\frac{\sqrt{57}}{6}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{\sqrt{57}}{6}-\frac{3}{2} x=-\frac{\sqrt{57}}{6}-\frac{3}{2}$$