Solve 0=3/256x^2-3/16x+5/12

$$0 = \frac{ 3 }{ 256 } { x }^{ 2 } - \frac{ 3 }{ 16 } x+ \frac{ 5 }{ 12 }$$

Solve for x

$$x = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

$$x = \frac{40}{3} = 13\frac{1}{3} \approx 13.333333333$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$\frac{3}{256}x^{2}-\frac{3}{16}x+\frac{5}{12}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{3}{256} for a, -\frac{3}{16} for b, and \frac{5}{12} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{3}{16}\right)±\sqrt{\left(-\frac{3}{16}\right)^{2}-4\times \frac{3}{256}\times \frac{5}{12}}}{2\times \frac{3}{256}}$$

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

$$x=\frac{-\left(-\frac{3}{16}\right)±\sqrt{\frac{9}{256}-4\times \frac{3}{256}\times \frac{5}{12}}}{2\times \frac{3}{256}}$$

Multiply -4 times \frac{3}{256}.

$$x=\frac{-\left(-\frac{3}{16}\right)±\sqrt{\frac{9}{256}-\frac{3}{64}\times \frac{5}{12}}}{2\times \frac{3}{256}}$$

Multiply -\frac{3}{64} times \frac{5}{12} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{3}{16}\right)±\sqrt{\frac{9-5}{256}}}{2\times \frac{3}{256}}$$

Add \frac{9}{256} to -\frac{5}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\left(-\frac{3}{16}\right)±\sqrt{\frac{1}{64}}}{2\times \frac{3}{256}}$$

Take the square root of \frac{1}{64}.

$$x=\frac{-\left(-\frac{3}{16}\right)±\frac{1}{8}}{2\times \frac{3}{256}}$$

The opposite of -\frac{3}{16} is \frac{3}{16}.

$$x=\frac{\frac{3}{16}±\frac{1}{8}}{2\times \frac{3}{256}}$$

Multiply 2 times \frac{3}{256}.

$$x=\frac{\frac{3}{16}±\frac{1}{8}}{\frac{3}{128}}$$

Now solve the equation x=\frac{\frac{3}{16}±\frac{1}{8}}{\frac{3}{128}} when ± is plus. Add \frac{3}{16} to \frac{1}{8} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{5}{16}}{\frac{3}{128}}$$

Divide \frac{5}{16} by \frac{3}{128} by multiplying \frac{5}{16} by the reciprocal of \frac{3}{128}.

$$x=\frac{40}{3}$$

Now solve the equation x=\frac{\frac{3}{16}±\frac{1}{8}}{\frac{3}{128}} when ± is minus. Subtract \frac{1}{8} from \frac{3}{16} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{1}{16}}{\frac{3}{128}}$$

Divide \frac{1}{16} by \frac{3}{128} by multiplying \frac{1}{16} by the reciprocal of \frac{3}{128}.

$$x=\frac{8}{3}$$

The equation is now solved.

$$x=\frac{40}{3} x=\frac{8}{3}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$\frac{3}{256}x^{2}-\frac{3}{16}x+\frac{5}{12}=0$$

Subtract \frac{5}{12} from both sides. Anything subtracted from zero gives its negation.

$$\frac{3}{256}x^{2}-\frac{3}{16}x=-\frac{5}{12}$$

Divide both sides of the equation by \frac{3}{256}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{3}{256}x^{2}-\frac{3}{16}x}{\frac{3}{256}}=-\frac{\frac{5}{12}}{\frac{3}{256}}$$

Dividing by \frac{3}{256} undoes the multiplication by \frac{3}{256}.

$$x^{2}+\left(-\frac{\frac{3}{16}}{\frac{3}{256}}\right)x=-\frac{\frac{5}{12}}{\frac{3}{256}}$$

Divide -\frac{3}{16} by \frac{3}{256} by multiplying -\frac{3}{16} by the reciprocal of \frac{3}{256}.

$$x^{2}-16x=-\frac{\frac{5}{12}}{\frac{3}{256}}$$

Divide -\frac{5}{12} by \frac{3}{256} by multiplying -\frac{5}{12} by the reciprocal of \frac{3}{256}.

$$x^{2}-16x=-\frac{320}{9}$$

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

$$x^{2}-16x+\left(-8\right)^{2}=-\frac{320}{9}+\left(-8\right)^{2}$$

Square -8.

$$x^{2}-16x+64=-\frac{320}{9}+64$$

Add -\frac{320}{9} to 64.

$$x^{2}-16x+64=\frac{256}{9}$$

Factor x^{2}-16x+64. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-8\right)^{2}=\frac{256}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-8\right)^{2}}=\sqrt{\frac{256}{9}}$$

Simplify.

$$x-8=\frac{16}{3} x-8=-\frac{16}{3}$$

Add 8 to both sides of the equation.

$$x=\frac{40}{3} x=\frac{8}{3}$$