Solve 3a^2-4a=8+6a

$$3 a ^ { 2 } - 4 a = 8 + 6 a$$

Solve for a

$$a=-\frac{2}{3}\approx -0.666666667$$

$$a=4$$

Steps Using Factoring By Grouping

Subtract 8 from both sides.

$$3a^{2}-4a-8=6a$$

Subtract 6a from both sides.

$$3a^{2}-4a-8-6a=0$$

Combine -4a and -6a to get -10a.

$$3a^{2}-10a-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3a^{2}+aa+ba-8. To find a and b, set up a system to be solved.

$$a+b=-10 ab=3\left(-8\right)=-24$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -24.

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

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -10.

$$a=-12 b=2$$

Rewrite 3a^{2}-10a-8 as \left(3a^{2}-12a\right)+\left(2a-8\right).

$$\left(3a^{2}-12a\right)+\left(2a-8\right)$$

Factor out 3a in the first and 2 in the second group.

$$3a\left(a-4\right)+2\left(a-4\right)$$

Factor out common term a-4 by using distributive property.

$$\left(a-4\right)\left(3a+2\right)$$

To find equation solutions, solve a-4=0 and 3a+2=0.

$$a=4 a=-\frac{2}{3}$$

Steps Using the Quadratic Formula

Subtract 8 from both sides.

$$3a^{2}-4a-8=6a$$

Subtract 6a from both sides.

$$3a^{2}-4a-8-6a=0$$

Combine -4a and -6a to get -10a.

$$3a^{2}-10a-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -10 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 3\left(-8\right)}}{2\times 3}$$

Square -10.

$$a=\frac{-\left(-10\right)±\sqrt{100-4\times 3\left(-8\right)}}{2\times 3}$$

Multiply -4 times 3.

$$a=\frac{-\left(-10\right)±\sqrt{100-12\left(-8\right)}}{2\times 3}$$

Multiply -12 times -8.

$$a=\frac{-\left(-10\right)±\sqrt{100+96}}{2\times 3}$$

Add 100 to 96.

$$a=\frac{-\left(-10\right)±\sqrt{196}}{2\times 3}$$

Take the square root of 196.

$$a=\frac{-\left(-10\right)±14}{2\times 3}$$

The opposite of -10 is 10.

$$a=\frac{10±14}{2\times 3}$$

Multiply 2 times 3.

$$a=\frac{10±14}{6}$$

Now solve the equation a=\frac{10±14}{6} when ± is plus. Add 10 to 14.

$$a=\frac{24}{6}$$

Divide 24 by 6.

$$a=4$$

Now solve the equation a=\frac{10±14}{6} when ± is minus. Subtract 14 from 10.

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

Reduce the fraction \frac{-4}{6} to lowest terms by extracting and canceling out 2.

$$a=-\frac{2}{3}$$

The equation is now solved.

$$a=4 a=-\frac{2}{3}$$

Steps for Completing the Square

Subtract 6a from both sides.

$$3a^{2}-4a-6a=8$$

Combine -4a and -6a to get -10a.

$$3a^{2}-10a=8$$

Divide both sides by 3.

$$\frac{3a^{2}-10a}{3}=\frac{8}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$a^{2}-\frac{10}{3}a=\frac{8}{3}$$

Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-\frac{10}{3}a+\left(-\frac{5}{3}\right)^{2}=\frac{8}{3}+\left(-\frac{5}{3}\right)^{2}$$

Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-\frac{10}{3}a+\frac{25}{9}=\frac{8}{3}+\frac{25}{9}$$

Add \frac{8}{3} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}-\frac{10}{3}a+\frac{25}{9}=\frac{49}{9}$$

Factor a^{2}-\frac{10}{3}a+\frac{25}{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}.

$$\left(a-\frac{5}{3}\right)^{2}=\frac{49}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{5}{3}\right)^{2}}=\sqrt{\frac{49}{9}}$$

Simplify.

$$a-\frac{5}{3}=\frac{7}{3} a-\frac{5}{3}=-\frac{7}{3}$$

Add \frac{5}{3} to both sides of the equation.

$$a=4 a=-\frac{2}{3}$$