Solve (x)=3x^2-x-4

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

Solve for x

$$x = \frac{\sqrt{13} + 1}{3} \approx 1.535183758$$

$$x=\frac{1-\sqrt{13}}{3}\approx -0.868517092$$

Steps Using the Quadratic Formula

Subtract 3x^{2} from both sides.

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

Add x to both sides.

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

Combine x and x to get 2x.

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

Add 4 to both sides.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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

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

Square 2.

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

Multiply -4 times -3.

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

Multiply 12 times 4.

$$x=\frac{-2±\sqrt{4+48}}{2\left(-3\right)}$$

Add 4 to 48.

$$x=\frac{-2±\sqrt{52}}{2\left(-3\right)}$$

Take the square root of 52.

$$x=\frac{-2±2\sqrt{13}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-2±2\sqrt{13}}{-6}$$

Now solve the equation x=\frac{-2±2\sqrt{13}}{-6} when ± is plus. Add -2 to 2\sqrt{13}.

$$x=\frac{2\sqrt{13}-2}{-6}$$

Divide -2+2\sqrt{13} by -6.

$$x=\frac{1-\sqrt{13}}{3}$$

Now solve the equation x=\frac{-2±2\sqrt{13}}{-6} when ± is minus. Subtract 2\sqrt{13} from -2.

$$x=\frac{-2\sqrt{13}-2}{-6}$$

Divide -2-2\sqrt{13} by -6.

$$x=\frac{\sqrt{13}+1}{3}$$

The equation is now solved.

$$x=\frac{1-\sqrt{13}}{3} x=\frac{\sqrt{13}+1}{3}$$

Steps for Completing the Square

Subtract 3x^{2} from both sides.

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

Add x to both sides.

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

Combine x and x to get 2x.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide 2 by -3.

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

Divide -4 by -3.

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

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

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

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{4}{3}+\frac{1}{9}$$

Add \frac{4}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{13}{9}$$

Factor x^{2}-\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{13}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{13}{9}}$$

Simplify.

$$x-\frac{1}{3}=\frac{\sqrt{13}}{3} x-\frac{1}{3}=-\frac{\sqrt{13}}{3}$$

Add \frac{1}{3} to both sides of the equation.

$$x=\frac{\sqrt{13}+1}{3} x=\frac{1-\sqrt{13}}{3}$$