Solve 3x^2=x+4

$$3 { x }^{ 2 } =x+4$$

Solve for x

$$x=-1$$

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Steps Using Factoring By Grouping

Subtract x from both sides.

$$3x^{2}-x=4$$

Subtract 4 from both sides.

$$3x^{2}-x-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=-1 ab=3\left(-4\right)=-12$$

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

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

Factor out x in 3x^{2}-4x.

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

Factor out common term 3x-4 by using distributive property.

$$\left(3x-4\right)\left(x+1\right)$$

To find equation solutions, solve 3x-4=0 and x+1=0.

$$x=\frac{4}{3} x=-1$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$3x^{2}-x=4$$

Subtract 4 from both sides.

$$3x^{2}-x-4=0$$

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

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

Multiply -4 times 3.

$$x=\frac{-\left(-1\right)±\sqrt{1-12\left(-4\right)}}{2\times 3}$$

Multiply -12 times -4.

$$x=\frac{-\left(-1\right)±\sqrt{1+48}}{2\times 3}$$

Add 1 to 48.

$$x=\frac{-\left(-1\right)±\sqrt{49}}{2\times 3}$$

Take the square root of 49.

$$x=\frac{-\left(-1\right)±7}{2\times 3}$$

The opposite of -1 is 1.

$$x=\frac{1±7}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{1±7}{6}$$

Now solve the equation x=\frac{1±7}{6} when ± is plus. Add 1 to 7.

$$x=\frac{8}{6}$$

Reduce the fraction \frac{8}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{4}{3}$$

Now solve the equation x=\frac{1±7}{6} when ± is minus. Subtract 7 from 1.

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

Divide -6 by 6.

$$x=-1$$

The equation is now solved.

$$x=\frac{4}{3} x=-1$$

Steps for Completing the Square

Subtract x from both sides.

$$3x^{2}-x=4$$

Divide both sides by 3.

$$\frac{3x^{2}-x}{3}=\frac{4}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{1}{3}x=\frac{4}{3}$$

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.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{4}{3}+\left(-\frac{1}{6}\right)^{2}$$

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

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{4}{3}+\frac{1}{36}$$

Add \frac{4}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{49}{36}$$

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}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{7}{6} x-\frac{1}{6}=-\frac{7}{6}$$

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

$$x=\frac{4}{3} x=-1$$