Solve 10=3x^2+x

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

Solve for x

$$x=-2$$

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

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

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

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 positive, the positive number has greater absolute value than the negative. 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.

$$a=-5 b=6$$

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

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

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

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

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

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

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

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

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

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

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}.

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

Square 1.

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

Multiply -4 times 3.

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

Multiply -12 times -10.

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

Add 1 to 120.

$$x=\frac{-1±\sqrt{121}}{2\times 3}$$

Take the square root of 121.

$$x=\frac{-1±11}{2\times 3}$$

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.

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

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

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

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

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

Divide -12 by 6.

$$x=-2$$

The equation is now solved.

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

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

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

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.

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

Subtract \frac{1}{6} from both sides of the equation.

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