Solve 6x^2=x+15

$$6 x ^ { 2 } = x + 15$$

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

Steps Using Factoring By Grouping

Subtract x from both sides.

$$6x^{2}-x=15$$

Subtract 15 from both sides.

$$6x^{2}-x-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=-1 ab=6\left(-15\right)=-90$$

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 -90.

$$1,-90 2,-45 3,-30 5,-18 6,-15 9,-10$$

Calculate the sum for each pair.

$$1-90=-89 2-45=-43 3-30=-27 5-18=-13 6-15=-9 9-10=-1$$

The solution is the pair that gives sum -1.

$$a=-10 b=9$$

Rewrite 6x^{2}-x-15 as \left(6x^{2}-10x\right)+\left(9x-15\right).

$$\left(6x^{2}-10x\right)+\left(9x-15\right)$$

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

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

Factor out common term 3x-5 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Subtract x from both sides.

$$6x^{2}-x=15$$

Subtract 15 from both sides.

$$6x^{2}-x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -1 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 6\left(-15\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-1\right)±\sqrt{1-24\left(-15\right)}}{2\times 6}$$

Multiply -24 times -15.

$$x=\frac{-\left(-1\right)±\sqrt{1+360}}{2\times 6}$$

Add 1 to 360.

$$x=\frac{-\left(-1\right)±\sqrt{361}}{2\times 6}$$

Take the square root of 361.

$$x=\frac{-\left(-1\right)±19}{2\times 6}$$

The opposite of -1 is 1.

$$x=\frac{1±19}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{1±19}{12}$$

Now solve the equation x=\frac{1±19}{12} when ± is plus. Add 1 to 19.

$$x=\frac{20}{12}$$

Reduce the fraction \frac{20}{12} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{3}$$

Now solve the equation x=\frac{1±19}{12} when ± is minus. Subtract 19 from 1.

$$x=-\frac{18}{12}$$

Reduce the fraction \frac{-18}{12} to lowest terms by extracting and canceling out 6.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract x from both sides.

$$6x^{2}-x=15$$

Divide both sides by 6.

$$\frac{6x^{2}-x}{6}=\frac{15}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}-\frac{1}{6}x=\frac{15}{6}$$

Reduce the fraction \frac{15}{6} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{1}{6}x=\frac{5}{2}$$

Divide -\frac{1}{6}, the coefficient of the x term, by 2 to get -\frac{1}{12}. Then add the square of -\frac{1}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{6}x+\left(-\frac{1}{12}\right)^{2}=\frac{5}{2}+\left(-\frac{1}{12}\right)^{2}$$

Square -\frac{1}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{6}x+\frac{1}{144}=\frac{5}{2}+\frac{1}{144}$$

Add \frac{5}{2} to \frac{1}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{6}x+\frac{1}{144}=\frac{361}{144}$$

Factor x^{2}-\frac{1}{6}x+\frac{1}{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}.

$$\left(x-\frac{1}{12}\right)^{2}=\frac{361}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{12}\right)^{2}}=\sqrt{\frac{361}{144}}$$

Simplify.

$$x-\frac{1}{12}=\frac{19}{12} x-\frac{1}{12}=-\frac{19}{12}$$

Add \frac{1}{12} to both sides of the equation.

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