Solve 8-4x+5x^2/12=0

$$8-4x+ \frac{ 5 { x }^{ 2 } }{ 12 } = 0$$

Solve for x

$$x = \frac{4 \sqrt{6} + 24}{5} \approx 6.759591794$$

$$x = \frac{24 - 4 \sqrt{6}}{5} \approx 2.840408206$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 12.

$$96-48x+5x^{2}=0$$

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.

$$5x^{2}-48x+96=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -48 for b, and 96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-48\right)±\sqrt{\left(-48\right)^{2}-4\times 5\times 96}}{2\times 5}$$

Square -48.

$$x=\frac{-\left(-48\right)±\sqrt{2304-4\times 5\times 96}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-48\right)±\sqrt{2304-20\times 96}}{2\times 5}$$

Multiply -20 times 96.

$$x=\frac{-\left(-48\right)±\sqrt{2304-1920}}{2\times 5}$$

Add 2304 to -1920.

$$x=\frac{-\left(-48\right)±\sqrt{384}}{2\times 5}$$

Take the square root of 384.

$$x=\frac{-\left(-48\right)±8\sqrt{6}}{2\times 5}$$

The opposite of -48 is 48.

$$x=\frac{48±8\sqrt{6}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{48±8\sqrt{6}}{10}$$

Now solve the equation x=\frac{48±8\sqrt{6}}{10} when ± is plus. Add 48 to 8\sqrt{6}.

$$x=\frac{8\sqrt{6}+48}{10}$$

Divide 48+8\sqrt{6} by 10.

$$x=\frac{4\sqrt{6}+24}{5}$$

Now solve the equation x=\frac{48±8\sqrt{6}}{10} when ± is minus. Subtract 8\sqrt{6} from 48.

$$x=\frac{48-8\sqrt{6}}{10}$$

Divide 48-8\sqrt{6} by 10.

$$x=\frac{24-4\sqrt{6}}{5}$$

The equation is now solved.

$$x=\frac{4\sqrt{6}+24}{5} x=\frac{24-4\sqrt{6}}{5}$$

Steps for Completing the Square

Multiply both sides of the equation by 12.

$$96-48x+5x^{2}=0$$

Subtract 96 from both sides. Anything subtracted from zero gives its negation.

$$-48x+5x^{2}=-96$$

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.

$$5x^{2}-48x=-96$$

Divide both sides by 5.

$$\frac{5x^{2}-48x}{5}=-\frac{96}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{48}{5}x=-\frac{96}{5}$$

Divide -\frac{48}{5}, the coefficient of the x term, by 2 to get -\frac{24}{5}. Then add the square of -\frac{24}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{48}{5}x+\left(-\frac{24}{5}\right)^{2}=-\frac{96}{5}+\left(-\frac{24}{5}\right)^{2}$$

Square -\frac{24}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{48}{5}x+\frac{576}{25}=-\frac{96}{5}+\frac{576}{25}$$

Add -\frac{96}{5} to \frac{576}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{48}{5}x+\frac{576}{25}=\frac{96}{25}$$

Factor x^{2}-\frac{48}{5}x+\frac{576}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{24}{5}\right)^{2}=\frac{96}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{24}{5}\right)^{2}}=\sqrt{\frac{96}{25}}$$

Simplify.

$$x-\frac{24}{5}=\frac{4\sqrt{6}}{5} x-\frac{24}{5}=-\frac{4\sqrt{6}}{5}$$

Add \frac{24}{5} to both sides of the equation.

$$x=\frac{4\sqrt{6}+24}{5} x=\frac{24-4\sqrt{6}}{5}$$