Solve -9x=2(3x^2+4)+10x

$$- 9 x = 2 ( 3 x ^ { 2 } + 4 ) + 10 x$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x = -\frac{8}{3} = -2\frac{2}{3} \approx -2.666666667$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 2 by 3x^{2}+4.

$$-9x=6x^{2}+8+10x$$

Subtract 6x^{2} from both sides.

$$-9x-6x^{2}=8+10x$$

Subtract 8 from both sides.

$$-9x-6x^{2}-8=10x$$

Subtract 10x from both sides.

$$-9x-6x^{2}-8-10x=0$$

Combine -9x and -10x to get -19x.

$$-19x-6x^{2}-8=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-6x^{2}-19x-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -6x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=-19 ab=-6\left(-8\right)=48$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 48.

$$-1,-48 -2,-24 -3,-16 -4,-12 -6,-8$$

Calculate the sum for each pair.

$$-1-48=-49 -2-24=-26 -3-16=-19 -4-12=-16 -6-8=-14$$

The solution is the pair that gives sum -19.

$$a=-3 b=-16$$

Rewrite -6x^{2}-19x-8 as \left(-6x^{2}-3x\right)+\left(-16x-8\right).

$$\left(-6x^{2}-3x\right)+\left(-16x-8\right)$$

Factor out -3x in the first and -8 in the second group.

$$-3x\left(2x+1\right)-8\left(2x+1\right)$$

Factor out common term 2x+1 by using distributive property.

$$\left(2x+1\right)\left(-3x-8\right)$$

To find equation solutions, solve 2x+1=0 and -3x-8=0.

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

Steps Using the Quadratic Formula

Use the distributive property to multiply 2 by 3x^{2}+4.

$$-9x=6x^{2}+8+10x$$

Subtract 6x^{2} from both sides.

$$-9x-6x^{2}=8+10x$$

Subtract 8 from both sides.

$$-9x-6x^{2}-8=10x$$

Subtract 10x from both sides.

$$-9x-6x^{2}-8-10x=0$$

Combine -9x and -10x to get -19x.

$$-19x-6x^{2}-8=0$$

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.

$$-6x^{2}-19x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, -19 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\left(-6\right)\left(-8\right)}}{2\left(-6\right)}$$

Square -19.

$$x=\frac{-\left(-19\right)±\sqrt{361-4\left(-6\right)\left(-8\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-\left(-19\right)±\sqrt{361+24\left(-8\right)}}{2\left(-6\right)}$$

Multiply 24 times -8.

$$x=\frac{-\left(-19\right)±\sqrt{361-192}}{2\left(-6\right)}$$

Add 361 to -192.

$$x=\frac{-\left(-19\right)±\sqrt{169}}{2\left(-6\right)}$$

Take the square root of 169.

$$x=\frac{-\left(-19\right)±13}{2\left(-6\right)}$$

The opposite of -19 is 19.

$$x=\frac{19±13}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{19±13}{-12}$$

Now solve the equation x=\frac{19±13}{-12} when ± is plus. Add 19 to 13.

$$x=\frac{32}{-12}$$

Reduce the fraction \frac{32}{-12} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{19±13}{-12} when ± is minus. Subtract 13 from 19.

$$x=\frac{6}{-12}$$

Reduce the fraction \frac{6}{-12} to lowest terms by extracting and canceling out 6.

$$x=-\frac{1}{2}$$

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply 2 by 3x^{2}+4.

$$-9x=6x^{2}+8+10x$$

Subtract 6x^{2} from both sides.

$$-9x-6x^{2}=8+10x$$

Subtract 10x from both sides.

$$-9x-6x^{2}-10x=8$$

Combine -9x and -10x to get -19x.

$$-19x-6x^{2}=8$$

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.

$$-6x^{2}-19x=8$$

Divide both sides by -6.

$$\frac{-6x^{2}-19x}{-6}=\frac{8}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\left(-\frac{19}{-6}\right)x=\frac{8}{-6}$$

Divide -19 by -6.

$$x^{2}+\frac{19}{6}x=\frac{8}{-6}$$

Reduce the fraction \frac{8}{-6} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{19}{6}x=-\frac{4}{3}$$

Divide \frac{19}{6}, the coefficient of the x term, by 2 to get \frac{19}{12}. Then add the square of \frac{19}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{19}{6}x+\left(\frac{19}{12}\right)^{2}=-\frac{4}{3}+\left(\frac{19}{12}\right)^{2}$$

Square \frac{19}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{19}{6}x+\frac{361}{144}=-\frac{4}{3}+\frac{361}{144}$$

Add -\frac{4}{3} to \frac{361}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{19}{6}x+\frac{361}{144}=\frac{169}{144}$$

Factor x^{2}+\frac{19}{6}x+\frac{361}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{19}{12}\right)^{2}=\frac{169}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{19}{12}\right)^{2}}=\sqrt{\frac{169}{144}}$$

Simplify.

$$x+\frac{19}{12}=\frac{13}{12} x+\frac{19}{12}=-\frac{13}{12}$$

Subtract \frac{19}{12} from both sides of the equation.

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