Solve 3x^2/2-11x-4/2=0

$$\frac { 3 x ^ { 2 } } { 2 } - \frac { 11 x - 4 } { 2 } = 0$$

Solve for x

$$x = \frac{\sqrt{73} + 11}{6} \approx 3.257333958$$

$$x=\frac{11-\sqrt{73}}{6}\approx 0.409332709$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$3x^{2}-\left(11x-4\right)=0$$

To find the opposite of 11x-4, find the opposite of each term.

$$3x^{2}-11x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -11 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 3\times 4}}{2\times 3}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 3\times 4}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-11\right)±\sqrt{121-12\times 4}}{2\times 3}$$

Multiply -12 times 4.

$$x=\frac{-\left(-11\right)±\sqrt{121-48}}{2\times 3}$$

Add 121 to -48.

$$x=\frac{-\left(-11\right)±\sqrt{73}}{2\times 3}$$

The opposite of -11 is 11.

$$x=\frac{11±\sqrt{73}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{11±\sqrt{73}}{6}$$

Now solve the equation x=\frac{11±\sqrt{73}}{6} when ± is plus. Add 11 to \sqrt{73}.

$$x=\frac{\sqrt{73}+11}{6}$$

Now solve the equation x=\frac{11±\sqrt{73}}{6} when ± is minus. Subtract \sqrt{73} from 11.

$$x=\frac{11-\sqrt{73}}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{73}+11}{6} x=\frac{11-\sqrt{73}}{6}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$3x^{2}-\left(11x-4\right)=0$$

To find the opposite of 11x-4, find the opposite of each term.

$$3x^{2}-11x+4=0$$

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

$$3x^{2}-11x=-4$$

Divide both sides by 3.

$$\frac{3x^{2}-11x}{3}=-\frac{4}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{11}{3}x=-\frac{4}{3}$$

Divide -\frac{11}{3}, the coefficient of the x term, by 2 to get -\frac{11}{6}. Then add the square of -\frac{11}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{3}x+\left(-\frac{11}{6}\right)^{2}=-\frac{4}{3}+\left(-\frac{11}{6}\right)^{2}$$

Square -\frac{11}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=-\frac{4}{3}+\frac{121}{36}$$

Add -\frac{4}{3} to \frac{121}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=\frac{73}{36}$$

Factor x^{2}-\frac{11}{3}x+\frac{121}{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}.

$$\left(x-\frac{11}{6}\right)^{2}=\frac{73}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{6}\right)^{2}}=\sqrt{\frac{73}{36}}$$

Simplify.

$$x-\frac{11}{6}=\frac{\sqrt{73}}{6} x-\frac{11}{6}=-\frac{\sqrt{73}}{6}$$

Add \frac{11}{6} to both sides of the equation.

$$x=\frac{\sqrt{73}+11}{6} x=\frac{11-\sqrt{73}}{6}$$