Solve 18+12x+6x+4x^2=31

$$18+12x+6x+4 { x }^{ 2 } =31$$

Solve for x

$$x=\frac{\sqrt{133}-9}{4}\approx 0.633140649$$

$$x=\frac{-\sqrt{133}-9}{4}\approx -5.133140649$$

Steps Using the Quadratic Formula

Combine 12x and 6x to get 18x.

$$18+18x+4x^{2}=31$$

Subtract 31 from both sides.

$$18+18x+4x^{2}-31=0$$

Subtract 31 from 18 to get -13.

$$-13+18x+4x^{2}=0$$

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.

$$4x^{2}+18x-13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 18 for b, and -13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\times 4\left(-13\right)}}{2\times 4}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\times 4\left(-13\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-18±\sqrt{324-16\left(-13\right)}}{2\times 4}$$

Multiply -16 times -13.

$$x=\frac{-18±\sqrt{324+208}}{2\times 4}$$

Add 324 to 208.

$$x=\frac{-18±\sqrt{532}}{2\times 4}$$

Take the square root of 532.

$$x=\frac{-18±2\sqrt{133}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-18±2\sqrt{133}}{8}$$

Now solve the equation x=\frac{-18±2\sqrt{133}}{8} when ± is plus. Add -18 to 2\sqrt{133}.

$$x=\frac{2\sqrt{133}-18}{8}$$

Divide -18+2\sqrt{133} by 8.

$$x=\frac{\sqrt{133}-9}{4}$$

Now solve the equation x=\frac{-18±2\sqrt{133}}{8} when ± is minus. Subtract 2\sqrt{133} from -18.

$$x=\frac{-2\sqrt{133}-18}{8}$$

Divide -18-2\sqrt{133} by 8.

$$x=\frac{-\sqrt{133}-9}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{133}-9}{4} x=\frac{-\sqrt{133}-9}{4}$$

Steps for Completing the Square

Combine 12x and 6x to get 18x.

$$18+18x+4x^{2}=31$$

Subtract 18 from both sides.

$$18x+4x^{2}=31-18$$

Subtract 18 from 31 to get 13.

$$18x+4x^{2}=13$$

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.

$$4x^{2}+18x=13$$

Divide both sides by 4.

$$\frac{4x^{2}+18x}{4}=\frac{13}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{18}{4}x=\frac{13}{4}$$

Reduce the fraction \frac{18}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{9}{2}x=\frac{13}{4}$$

Divide \frac{9}{2}, the coefficient of the x term, by 2 to get \frac{9}{4}. Then add the square of \frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{9}{2}x+\frac{81}{16}=\frac{13}{4}+\frac{81}{16}$$

Add \frac{13}{4} to \frac{81}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{9}{2}x+\frac{81}{16}=\frac{133}{16}$$

Factor x^{2}+\frac{9}{2}x+\frac{81}{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}.

$$\left(x+\frac{9}{4}\right)^{2}=\frac{133}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{4}\right)^{2}}=\sqrt{\frac{133}{16}}$$

Simplify.

$$x+\frac{9}{4}=\frac{\sqrt{133}}{4} x+\frac{9}{4}=-\frac{\sqrt{133}}{4}$$

Subtract \frac{9}{4} from both sides of the equation.

$$x=\frac{\sqrt{133}-9}{4} x=\frac{-\sqrt{133}-9}{4}$$