Solve 2x^2/3+x/4=5

$$\frac { 2 x ^ { 2 } } { 3 } + \frac { x } { 4 } = 5$$

Solve for x

$$x = \frac{\sqrt{1929} - 3}{16} \approx 2.557523907$$

$$x=\frac{-\sqrt{1929}-3}{16}\approx -2.932523907$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 12, the least common multiple of 3,4.

$$4\times 2x^{2}+3x=60$$

Multiply 4 and 2 to get 8.

$$8x^{2}+3x=60$$

Subtract 60 from both sides.

$$8x^{2}+3x-60=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 3 for b, and -60 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times 8\left(-60\right)}}{2\times 8}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 8\left(-60\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-3±\sqrt{9-32\left(-60\right)}}{2\times 8}$$

Multiply -32 times -60.

$$x=\frac{-3±\sqrt{9+1920}}{2\times 8}$$

Add 9 to 1920.

$$x=\frac{-3±\sqrt{1929}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-3±\sqrt{1929}}{16}$$

Now solve the equation x=\frac{-3±\sqrt{1929}}{16} when ± is plus. Add -3 to \sqrt{1929}.

$$x=\frac{\sqrt{1929}-3}{16}$$

Now solve the equation x=\frac{-3±\sqrt{1929}}{16} when ± is minus. Subtract \sqrt{1929} from -3.

$$x=\frac{-\sqrt{1929}-3}{16}$$

The equation is now solved.

$$x=\frac{\sqrt{1929}-3}{16} x=\frac{-\sqrt{1929}-3}{16}$$

Steps for Completing the Square

Multiply both sides of the equation by 12, the least common multiple of 3,4.

$$4\times 2x^{2}+3x=60$$

Multiply 4 and 2 to get 8.

$$8x^{2}+3x=60$$

Divide both sides by 8.

$$\frac{8x^{2}+3x}{8}=\frac{60}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\frac{3}{8}x=\frac{60}{8}$$

Reduce the fraction \frac{60}{8} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{3}{8}x=\frac{15}{2}$$

Divide \frac{3}{8}, the coefficient of the x term, by 2 to get \frac{3}{16}. Then add the square of \frac{3}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{8}x+\left(\frac{3}{16}\right)^{2}=\frac{15}{2}+\left(\frac{3}{16}\right)^{2}$$

Square \frac{3}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{8}x+\frac{9}{256}=\frac{15}{2}+\frac{9}{256}$$

Add \frac{15}{2} to \frac{9}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{8}x+\frac{9}{256}=\frac{1929}{256}$$

Factor x^{2}+\frac{3}{8}x+\frac{9}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{16}\right)^{2}=\frac{1929}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{16}\right)^{2}}=\sqrt{\frac{1929}{256}}$$

Simplify.

$$x+\frac{3}{16}=\frac{\sqrt{1929}}{16} x+\frac{3}{16}=-\frac{\sqrt{1929}}{16}$$

Subtract \frac{3}{16} from both sides of the equation.

$$x=\frac{\sqrt{1929}-3}{16} x=\frac{-\sqrt{1929}-3}{16}$$