Solve 15x^2-12=8x

$$15 { x }^{ 2 } -12=8x$$

Solve for x

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

$$x = \frac{6}{5} = 1\frac{1}{5} = 1.2$$

Steps Using Factoring By Grouping

Subtract 8x from both sides.

$$15x^{2}-12-8x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$15x^{2}-8x-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 15x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=-8 ab=15\left(-12\right)=-180$$

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 -180.

$$1,-180 2,-90 3,-60 4,-45 5,-36 6,-30 9,-20 10,-18 12,-15$$

Calculate the sum for each pair.

$$1-180=-179 2-90=-88 3-60=-57 4-45=-41 5-36=-31 6-30=-24 9-20=-11 10-18=-8 12-15=-3$$

The solution is the pair that gives sum -8.

$$a=-18 b=10$$

Rewrite 15x^{2}-8x-12 as \left(15x^{2}-18x\right)+\left(10x-12\right).

$$\left(15x^{2}-18x\right)+\left(10x-12\right)$$

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

$$3x\left(5x-6\right)+2\left(5x-6\right)$$

Factor out common term 5x-6 by using distributive property.

$$\left(5x-6\right)\left(3x+2\right)$$

To find equation solutions, solve 5x-6=0 and 3x+2=0.

$$x=\frac{6}{5} x=-\frac{2}{3}$$

Steps Using the Quadratic Formula

Subtract 8x from both sides.

$$15x^{2}-12-8x=0$$

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.

$$15x^{2}-8x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 15 for a, -8 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 15\left(-12\right)}}{2\times 15}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 15\left(-12\right)}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-\left(-8\right)±\sqrt{64-60\left(-12\right)}}{2\times 15}$$

Multiply -60 times -12.

$$x=\frac{-\left(-8\right)±\sqrt{64+720}}{2\times 15}$$

Add 64 to 720.

$$x=\frac{-\left(-8\right)±\sqrt{784}}{2\times 15}$$

Take the square root of 784.

$$x=\frac{-\left(-8\right)±28}{2\times 15}$$

The opposite of -8 is 8.

$$x=\frac{8±28}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{8±28}{30}$$

Now solve the equation x=\frac{8±28}{30} when ± is plus. Add 8 to 28.

$$x=\frac{36}{30}$$

Reduce the fraction \frac{36}{30} to lowest terms by extracting and canceling out 6.

$$x=\frac{6}{5}$$

Now solve the equation x=\frac{8±28}{30} when ± is minus. Subtract 28 from 8.

$$x=-\frac{20}{30}$$

Reduce the fraction \frac{-20}{30} to lowest terms by extracting and canceling out 10.

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

The equation is now solved.

$$x=\frac{6}{5} x=-\frac{2}{3}$$

Steps for Completing the Square

Subtract 8x from both sides.

$$15x^{2}-12-8x=0$$

Add 12 to both sides. Anything plus zero gives itself.

$$15x^{2}-8x=12$$

Divide both sides by 15.

$$\frac{15x^{2}-8x}{15}=\frac{12}{15}$$

Dividing by 15 undoes the multiplication by 15.

$$x^{2}-\frac{8}{15}x=\frac{12}{15}$$

Reduce the fraction \frac{12}{15} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{8}{15}x=\frac{4}{5}$$

Divide -\frac{8}{15}, the coefficient of the x term, by 2 to get -\frac{4}{15}. Then add the square of -\frac{4}{15} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{15}x+\left(-\frac{4}{15}\right)^{2}=\frac{4}{5}+\left(-\frac{4}{15}\right)^{2}$$

Square -\frac{4}{15} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{15}x+\frac{16}{225}=\frac{4}{5}+\frac{16}{225}$$

Add \frac{4}{5} to \frac{16}{225} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{8}{15}x+\frac{16}{225}=\frac{196}{225}$$

Factor x^{2}-\frac{8}{15}x+\frac{16}{225}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{4}{15}\right)^{2}=\frac{196}{225}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{15}\right)^{2}}=\sqrt{\frac{196}{225}}$$

Simplify.

$$x-\frac{4}{15}=\frac{14}{15} x-\frac{4}{15}=-\frac{14}{15}$$

Add \frac{4}{15} to both sides of the equation.

$$x=\frac{6}{5} x=-\frac{2}{3}$$