Solve x^2-x/90=2/15

$$\frac { x ^ { 2 } - x } { 90 } = \frac { 2 } { 15 }$$

Solve for x

$$x=-3$$

$$x=4$$

Steps Using Factoring

Multiply both sides by 90.

$$x^{2}-x=\frac{2}{15}\times 90$$

Multiply \frac{2}{15} and 90 to get 12.

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

Subtract 12 from both sides.

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

To solve the equation, factor x^{2}-x-12 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-1 ab=-12$$

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 -12.

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

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -1.

$$a=-4 b=3$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

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

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

$$x=4 x=-3$$

Steps Using Factoring By Grouping

Multiply both sides by 90.

$$x^{2}-x=\frac{2}{15}\times 90$$

Multiply \frac{2}{15} and 90 to get 12.

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

Subtract 12 from both sides.

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

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

$$a+b=-1 ab=1\left(-12\right)=-12$$

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 -12.

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

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -1.

$$a=-4 b=3$$

Rewrite x^{2}-x-12 as \left(x^{2}-4x\right)+\left(3x-12\right).

$$\left(x^{2}-4x\right)+\left(3x-12\right)$$

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

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

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

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

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

$$x=4 x=-3$$

Steps Using the Quadratic Formula

Multiply both sides by 90.

$$x^{2}-x=\frac{2}{15}\times 90$$

Multiply \frac{2}{15} and 90 to get 12.

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

Subtract 12 from both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-12\right)}}{2}$$

Multiply -4 times -12.

$$x=\frac{-\left(-1\right)±\sqrt{1+48}}{2}$$

Add 1 to 48.

$$x=\frac{-\left(-1\right)±\sqrt{49}}{2}$$

Take the square root of 49.

$$x=\frac{-\left(-1\right)±7}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±7}{2}$$

Now solve the equation x=\frac{1±7}{2} when ± is plus. Add 1 to 7.

$$x=\frac{8}{2}$$

Divide 8 by 2.

$$x=4$$

Now solve the equation x=\frac{1±7}{2} when ± is minus. Subtract 7 from 1.

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

Divide -6 by 2.

$$x=-3$$

The equation is now solved.

$$x=4 x=-3$$

Steps for Completing the Square

Multiply both sides by 90.

$$x^{2}-x=\frac{2}{15}\times 90$$

Multiply \frac{2}{15} and 90 to get 12.

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

Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=12+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=12+\frac{1}{4}$$

Add 12 to \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{49}{4}$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x-\frac{1}{2}=\frac{7}{2} x-\frac{1}{2}=-\frac{7}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$x=4 x=-3$$