Solve 3x^2=x+10

$$3 x ^ { 2 } = x + 10$$

Solve for x

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

$$x=2$$

Steps Using Factoring By Grouping

Subtract x from both sides.

$$3x^{2}-x=10$$

Subtract 10 from both sides.

$$3x^{2}-x-10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-10. To find a and b, set up a system to be solved.

$$a+b=-1 ab=3\left(-10\right)=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -1.

$$a=-6 b=5$$

Rewrite 3x^{2}-x-10 as \left(3x^{2}-6x\right)+\left(5x-10\right).

$$\left(3x^{2}-6x\right)+\left(5x-10\right)$$

Factor out 3x in the first and 5 in the second group.

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

Factor out common term x-2 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Subtract x from both sides.

$$3x^{2}-x=10$$

Subtract 10 from both sides.

$$3x^{2}-x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -1 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 3\left(-10\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-1\right)±\sqrt{1-12\left(-10\right)}}{2\times 3}$$

Multiply -12 times -10.

$$x=\frac{-\left(-1\right)±\sqrt{1+120}}{2\times 3}$$

Add 1 to 120.

$$x=\frac{-\left(-1\right)±\sqrt{121}}{2\times 3}$$

Take the square root of 121.

$$x=\frac{-\left(-1\right)±11}{2\times 3}$$

The opposite of -1 is 1.

$$x=\frac{1±11}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{1±11}{6}$$

Now solve the equation x=\frac{1±11}{6} when ± is plus. Add 1 to 11.

$$x=\frac{12}{6}$$

Divide 12 by 6.

$$x=2$$

Now solve the equation x=\frac{1±11}{6} when ± is minus. Subtract 11 from 1.

$$x=-\frac{10}{6}$$

Reduce the fraction \frac{-10}{6} to lowest terms by extracting and canceling out 2.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract x from both sides.

$$3x^{2}-x=10$$

Divide both sides by 3.

$$\frac{3x^{2}-x}{3}=\frac{10}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{1}{3}x=\frac{10}{3}$$

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{10}{3}+\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{10}{3}+\frac{1}{36}$$

Add \frac{10}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{121}{36}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{121}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{121}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{11}{6} x-\frac{1}{6}=-\frac{11}{6}$$

Add \frac{1}{6} to both sides of the equation.

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