Solve (125-3/4x)x=4800

$$(125- \frac{ 3 }{ 4 } x)x=4800$$

Solve for x

$$x=60$$

$$x = \frac{320}{3} = 106\frac{2}{3} \approx 106.666666667$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 125-\frac{3}{4}x by x.

$$125x-\frac{3}{4}xx=4800$$

Multiply x and x to get x^{2}.

$$125x-\frac{3}{4}x^{2}=4800$$

Subtract 4800 from both sides.

$$125x-\frac{3}{4}x^{2}-4800=0$$

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.

$$-\frac{3}{4}x^{2}+125x-4800=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{3}{4} for a, 125 for b, and -4800 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-125±\sqrt{125^{2}-4\left(-\frac{3}{4}\right)\left(-4800\right)}}{2\left(-\frac{3}{4}\right)}$$

Square 125.

$$x=\frac{-125±\sqrt{15625-4\left(-\frac{3}{4}\right)\left(-4800\right)}}{2\left(-\frac{3}{4}\right)}$$

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

$$x=\frac{-125±\sqrt{15625+3\left(-4800\right)}}{2\left(-\frac{3}{4}\right)}$$

Multiply 3 times -4800.

$$x=\frac{-125±\sqrt{15625-14400}}{2\left(-\frac{3}{4}\right)}$$

Add 15625 to -14400.

$$x=\frac{-125±\sqrt{1225}}{2\left(-\frac{3}{4}\right)}$$

Take the square root of 1225.

$$x=\frac{-125±35}{2\left(-\frac{3}{4}\right)}$$

Multiply 2 times -\frac{3}{4}.

$$x=\frac{-125±35}{-\frac{3}{2}}$$

Now solve the equation x=\frac{-125±35}{-\frac{3}{2}} when ± is plus. Add -125 to 35.

$$x=-\frac{90}{-\frac{3}{2}}$$

Divide -90 by -\frac{3}{2} by multiplying -90 by the reciprocal of -\frac{3}{2}.

$$x=60$$

Now solve the equation x=\frac{-125±35}{-\frac{3}{2}} when ± is minus. Subtract 35 from -125.

$$x=-\frac{160}{-\frac{3}{2}}$$

Divide -160 by -\frac{3}{2} by multiplying -160 by the reciprocal of -\frac{3}{2}.

$$x=\frac{320}{3}$$

The equation is now solved.

$$x=60 x=\frac{320}{3}$$

Steps for Completing the Square

Use the distributive property to multiply 125-\frac{3}{4}x by x.

$$125x-\frac{3}{4}xx=4800$$

Multiply x and x to get x^{2}.

$$125x-\frac{3}{4}x^{2}=4800$$

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.

$$-\frac{3}{4}x^{2}+125x=4800$$

Divide both sides of the equation by -\frac{3}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{3}{4}x^{2}+125x}{-\frac{3}{4}}=\frac{4800}{-\frac{3}{4}}$$

Dividing by -\frac{3}{4} undoes the multiplication by -\frac{3}{4}.

$$x^{2}+\frac{125}{-\frac{3}{4}}x=\frac{4800}{-\frac{3}{4}}$$

Divide 125 by -\frac{3}{4} by multiplying 125 by the reciprocal of -\frac{3}{4}.

$$x^{2}-\frac{500}{3}x=\frac{4800}{-\frac{3}{4}}$$

Divide 4800 by -\frac{3}{4} by multiplying 4800 by the reciprocal of -\frac{3}{4}.

$$x^{2}-\frac{500}{3}x=-6400$$

Divide -\frac{500}{3}, the coefficient of the x term, by 2 to get -\frac{250}{3}. Then add the square of -\frac{250}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{500}{3}x+\left(-\frac{250}{3}\right)^{2}=-6400+\left(-\frac{250}{3}\right)^{2}$$

Square -\frac{250}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{500}{3}x+\frac{62500}{9}=-6400+\frac{62500}{9}$$

Add -6400 to \frac{62500}{9}.

$$x^{2}-\frac{500}{3}x+\frac{62500}{9}=\frac{4900}{9}$$

Factor x^{2}-\frac{500}{3}x+\frac{62500}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{250}{3}\right)^{2}=\frac{4900}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{250}{3}\right)^{2}}=\sqrt{\frac{4900}{9}}$$

Simplify.

$$x-\frac{250}{3}=\frac{70}{3} x-\frac{250}{3}=-\frac{70}{3}$$

Add \frac{250}{3} to both sides of the equation.

$$x=\frac{320}{3} x=60$$