Solve 18x^2+42x=-24

$$18 x ^ { 2 } + 42 x = - 24$$

Solve for x

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

$$x=-1$$

Steps Using Factoring By Grouping

Add 24 to both sides.

$$18x^{2}+42x+24=0$$

Divide both sides by 6.

$$3x^{2}+7x+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=7 ab=3\times 4=12$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 7.

$$a=3 b=4$$

Rewrite 3x^{2}+7x+4 as \left(3x^{2}+3x\right)+\left(4x+4\right).

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

Factor out 3x in the first and 4 in the second group.

$$3x\left(x+1\right)+4\left(x+1\right)$$

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

$$\left(x+1\right)\left(3x+4\right)$$

To find equation solutions, solve x+1=0 and 3x+4=0.

$$x=-1 x=-\frac{4}{3}$$

Steps Using the Quadratic Formula

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.

$$18x^{2}+42x=-24$$

Add 24 to both sides of the equation.

$$18x^{2}+42x-\left(-24\right)=-24-\left(-24\right)$$

Subtracting -24 from itself leaves 0.

$$18x^{2}+42x-\left(-24\right)=0$$

Subtract -24 from 0.

$$18x^{2}+42x+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 18 for a, 42 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-42±\sqrt{42^{2}-4\times 18\times 24}}{2\times 18}$$

Square 42.

$$x=\frac{-42±\sqrt{1764-4\times 18\times 24}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-42±\sqrt{1764-72\times 24}}{2\times 18}$$

Multiply -72 times 24.

$$x=\frac{-42±\sqrt{1764-1728}}{2\times 18}$$

Add 1764 to -1728.

$$x=\frac{-42±\sqrt{36}}{2\times 18}$$

Take the square root of 36.

$$x=\frac{-42±6}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{-42±6}{36}$$

Now solve the equation x=\frac{-42±6}{36} when ± is plus. Add -42 to 6.

$$x=-\frac{36}{36}$$

Divide -36 by 36.

$$x=-1$$

Now solve the equation x=\frac{-42±6}{36} when ± is minus. Subtract 6 from -42.

$$x=-\frac{48}{36}$$

Reduce the fraction \frac{-48}{36} to lowest terms by extracting and canceling out 12.

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

The equation is now solved.

$$x=-1 x=-\frac{4}{3}$$

Steps for Completing the Square

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.

$$18x^{2}+42x=-24$$

Divide both sides by 18.

$$\frac{18x^{2}+42x}{18}=-\frac{24}{18}$$

Dividing by 18 undoes the multiplication by 18.

$$x^{2}+\frac{42}{18}x=-\frac{24}{18}$$

Reduce the fraction \frac{42}{18} to lowest terms by extracting and canceling out 6.

$$x^{2}+\frac{7}{3}x=-\frac{24}{18}$$

Reduce the fraction \frac{-24}{18} to lowest terms by extracting and canceling out 6.

$$x^{2}+\frac{7}{3}x=-\frac{4}{3}$$

Divide \frac{7}{3}, the coefficient of the x term, by 2 to get \frac{7}{6}. Then add the square of \frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{3}x+\left(\frac{7}{6}\right)^{2}=-\frac{4}{3}+\left(\frac{7}{6}\right)^{2}$$

Square \frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=-\frac{4}{3}+\frac{49}{36}$$

Add -\frac{4}{3} to \frac{49}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=\frac{1}{36}$$

Factor x^{2}+\frac{7}{3}x+\frac{49}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$x+\frac{7}{6}=\frac{1}{6} x+\frac{7}{6}=-\frac{1}{6}$$

Subtract \frac{7}{6} from both sides of the equation.

$$x=-1 x=-\frac{4}{3}$$