Solve 2x+3xx+9=8-6x

$$2x+3xx+9 = 8-6x$$

Solve for x

$$x=\frac{\sqrt{13}-4}{3}\approx -0.131482908$$

$$x=\frac{-\sqrt{13}-4}{3}\approx -2.535183758$$

Steps Using the Quadratic Formula

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

$$2x+3x^{2}+9=8-6x$$

Subtract 8 from both sides.

$$2x+3x^{2}+9-8=-6x$$

Subtract 8 from 9 to get 1.

$$2x+3x^{2}+1=-6x$$

Add 6x to both sides.

$$2x+3x^{2}+1+6x=0$$

Combine 2x and 6x to get 8x.

$$8x+3x^{2}+1=0$$

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.

$$3x^{2}+8x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 8 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 3}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-8±\sqrt{64-12}}{2\times 3}$$

Add 64 to -12.

$$x=\frac{-8±\sqrt{52}}{2\times 3}$$

Take the square root of 52.

$$x=\frac{-8±2\sqrt{13}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-8±2\sqrt{13}}{6}$$

Now solve the equation x=\frac{-8±2\sqrt{13}}{6} when ± is plus. Add -8 to 2\sqrt{13}.

$$x=\frac{2\sqrt{13}-8}{6}$$

Divide -8+2\sqrt{13} by 6.

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

Now solve the equation x=\frac{-8±2\sqrt{13}}{6} when ± is minus. Subtract 2\sqrt{13} from -8.

$$x=\frac{-2\sqrt{13}-8}{6}$$

Divide -8-2\sqrt{13} by 6.

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

The equation is now solved.

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

Steps for Completing the Square

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

$$2x+3x^{2}+9=8-6x$$

Add 6x to both sides.

$$2x+3x^{2}+9+6x=8$$

Combine 2x and 6x to get 8x.

$$8x+3x^{2}+9=8$$

Subtract 9 from both sides.

$$8x+3x^{2}=8-9$$

Subtract 9 from 8 to get -1.

$$8x+3x^{2}=-1$$

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.

$$3x^{2}+8x=-1$$

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide \frac{8}{3}, the coefficient of the x term, by 2 to get \frac{4}{3}. Then add the square of \frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{8}{3}x+\left(\frac{4}{3}\right)^{2}=-\frac{1}{3}+\left(\frac{4}{3}\right)^{2}$$

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

$$x^{2}+\frac{8}{3}x+\frac{16}{9}=-\frac{1}{3}+\frac{16}{9}$$

Add -\frac{1}{3} to \frac{16}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{8}{3}x+\frac{16}{9}=\frac{13}{9}$$

Factor x^{2}+\frac{8}{3}x+\frac{16}{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}.

$$\left(x+\frac{4}{3}\right)^{2}=\frac{13}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{4}{3}\right)^{2}}=\sqrt{\frac{13}{9}}$$

Simplify.

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

Subtract \frac{4}{3} from both sides of the equation.

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