Solve 18a^2+10a=-11a+4

$$18 a ^ { 2 } + 10 a = - 11 a + 4$$

Solve for a

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

$$a=\frac{1}{6}\approx 0.166666667$$

Steps Using Factoring By Grouping

Add 11a to both sides.

$$18a^{2}+10a+11a=4$$

Combine 10a and 11a to get 21a.

$$18a^{2}+21a=4$$

Subtract 4 from both sides.

$$18a^{2}+21a-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 18a^{2}+aa+ba-4. To find a and b, set up a system to be solved.

$$a+b=21 ab=18\left(-4\right)=-72$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -72.

$$-1,72 -2,36 -3,24 -4,18 -6,12 -8,9$$

Calculate the sum for each pair.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

The solution is the pair that gives sum 21.

$$a=-3 b=24$$

Rewrite 18a^{2}+21a-4 as \left(18a^{2}-3a\right)+\left(24a-4\right).

$$\left(18a^{2}-3a\right)+\left(24a-4\right)$$

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

$$3a\left(6a-1\right)+4\left(6a-1\right)$$

Factor out common term 6a-1 by using distributive property.

$$\left(6a-1\right)\left(3a+4\right)$$

To find equation solutions, solve 6a-1=0 and 3a+4=0.

$$a=\frac{1}{6} a=-\frac{4}{3}$$

Steps Using the Quadratic Formula

Add 11a to both sides.

$$18a^{2}+10a+11a=4$$

Combine 10a and 11a to get 21a.

$$18a^{2}+21a=4$$

Subtract 4 from both sides.

$$18a^{2}+21a-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 18 for a, 21 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-21±\sqrt{21^{2}-4\times 18\left(-4\right)}}{2\times 18}$$

Square 21.

$$a=\frac{-21±\sqrt{441-4\times 18\left(-4\right)}}{2\times 18}$$

Multiply -4 times 18.

$$a=\frac{-21±\sqrt{441-72\left(-4\right)}}{2\times 18}$$

Multiply -72 times -4.

$$a=\frac{-21±\sqrt{441+288}}{2\times 18}$$

Add 441 to 288.

$$a=\frac{-21±\sqrt{729}}{2\times 18}$$

Take the square root of 729.

$$a=\frac{-21±27}{2\times 18}$$

Multiply 2 times 18.

$$a=\frac{-21±27}{36}$$

Now solve the equation a=\frac{-21±27}{36} when ± is plus. Add -21 to 27.

$$a=\frac{6}{36}$$

Reduce the fraction \frac{6}{36} to lowest terms by extracting and canceling out 6.

$$a=\frac{1}{6}$$

Now solve the equation a=\frac{-21±27}{36} when ± is minus. Subtract 27 from -21.

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

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

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

The equation is now solved.

$$a=\frac{1}{6} a=-\frac{4}{3}$$

Steps for Completing the Square

Add 11a to both sides.

$$18a^{2}+10a+11a=4$$

Combine 10a and 11a to get 21a.

$$18a^{2}+21a=4$$

Divide both sides by 18.

$$\frac{18a^{2}+21a}{18}=\frac{4}{18}$$

Dividing by 18 undoes the multiplication by 18.

$$a^{2}+\frac{21}{18}a=\frac{4}{18}$$

Reduce the fraction \frac{21}{18} to lowest terms by extracting and canceling out 3.

$$a^{2}+\frac{7}{6}a=\frac{4}{18}$$

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

$$a^{2}+\frac{7}{6}a=\frac{2}{9}$$

Divide \frac{7}{6}, the coefficient of the x term, by 2 to get \frac{7}{12}. Then add the square of \frac{7}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+\frac{7}{6}a+\left(\frac{7}{12}\right)^{2}=\frac{2}{9}+\left(\frac{7}{12}\right)^{2}$$

Square \frac{7}{12} by squaring both the numerator and the denominator of the fraction.

$$a^{2}+\frac{7}{6}a+\frac{49}{144}=\frac{2}{9}+\frac{49}{144}$$

Add \frac{2}{9} to \frac{49}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}+\frac{7}{6}a+\frac{49}{144}=\frac{9}{16}$$

Factor a^{2}+\frac{7}{6}a+\frac{49}{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}.

$$\left(a+\frac{7}{12}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{7}{12}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$a+\frac{7}{12}=\frac{3}{4} a+\frac{7}{12}=-\frac{3}{4}$$

Subtract \frac{7}{12} from both sides of the equation.

$$a=\frac{1}{6} a=-\frac{4}{3}$$