Solve (x)=3x^2-4/6+x

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

Solve for x

$$x = \frac{\sqrt{17} + 3}{2} \approx 3.561552813$$

$$x=\frac{3-\sqrt{17}}{2}\approx -0.561552813$$

Steps Using the Quadratic Formula

Subtract \frac{3x^{2}-4}{6+x} from both sides.

$$x-\frac{3x^{2}-4}{6+x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{6+x}{6+x}.

$$\frac{x\left(6+x\right)}{6+x}-\frac{3x^{2}-4}{6+x}=0$$

Since \frac{x\left(6+x\right)}{6+x} and \frac{3x^{2}-4}{6+x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(6+x\right)-\left(3x^{2}-4\right)}{6+x}=0$$

Do the multiplications in x\left(6+x\right)-\left(3x^{2}-4\right).

$$\frac{6x+x^{2}-3x^{2}+4}{6+x}=0$$

Combine like terms in 6x+x^{2}-3x^{2}+4.

$$\frac{6x-2x^{2}+4}{6+x}=0$$

Variable x cannot be equal to -6 since division by zero is not defined. Multiply both sides of the equation by x+6.

$$6x-2x^{2}+4=0$$

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.

$$-2x^{2}+6x+4=0$$

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

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

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-2\right)\times 4}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-6±\sqrt{36+8\times 4}}{2\left(-2\right)}$$

Multiply 8 times 4.

$$x=\frac{-6±\sqrt{36+32}}{2\left(-2\right)}$$

Add 36 to 32.

$$x=\frac{-6±\sqrt{68}}{2\left(-2\right)}$$

Take the square root of 68.

$$x=\frac{-6±2\sqrt{17}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-6±2\sqrt{17}}{-4}$$

Now solve the equation x=\frac{-6±2\sqrt{17}}{-4} when ± is plus. Add -6 to 2\sqrt{17}.

$$x=\frac{2\sqrt{17}-6}{-4}$$

Divide -6+2\sqrt{17} by -4.

$$x=\frac{3-\sqrt{17}}{2}$$

Now solve the equation x=\frac{-6±2\sqrt{17}}{-4} when ± is minus. Subtract 2\sqrt{17} from -6.

$$x=\frac{-2\sqrt{17}-6}{-4}$$

Divide -6-2\sqrt{17} by -4.

$$x=\frac{\sqrt{17}+3}{2}$$

The equation is now solved.

$$x=\frac{3-\sqrt{17}}{2} x=\frac{\sqrt{17}+3}{2}$$

Steps for Completing the Square

Subtract \frac{3x^{2}-4}{6+x} from both sides.

$$x-\frac{3x^{2}-4}{6+x}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{6+x}{6+x}.

$$\frac{x\left(6+x\right)}{6+x}-\frac{3x^{2}-4}{6+x}=0$$

Since \frac{x\left(6+x\right)}{6+x} and \frac{3x^{2}-4}{6+x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x\left(6+x\right)-\left(3x^{2}-4\right)}{6+x}=0$$

Do the multiplications in x\left(6+x\right)-\left(3x^{2}-4\right).

$$\frac{6x+x^{2}-3x^{2}+4}{6+x}=0$$

Combine like terms in 6x+x^{2}-3x^{2}+4.

$$\frac{6x-2x^{2}+4}{6+x}=0$$

Variable x cannot be equal to -6 since division by zero is not defined. Multiply both sides of the equation by x+6.

$$6x-2x^{2}+4=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$6x-2x^{2}=-4$$

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.

$$-2x^{2}+6x=-4$$

Divide both sides by -2.

$$\frac{-2x^{2}+6x}{-2}=-\frac{4}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{6}{-2}x=-\frac{4}{-2}$$

Divide 6 by -2.

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

Divide -4 by -2.

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

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

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

Add 2 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{17}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{17}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{17}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{17}}{2} x-\frac{3}{2}=-\frac{\sqrt{17}}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{\sqrt{17}+3}{2} x=\frac{3-\sqrt{17}}{2}$$