Solve -5x+900/x=15

$$-5x+ \frac{ 900 }{ x } =15$$

Solve for x

$$x=-15$$

$$x=12$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$-5xx+900=15x$$

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

$$-5x^{2}+900=15x$$

Subtract 15x from both sides.

$$-5x^{2}+900-15x=0$$

Divide both sides by 5.

$$-x^{2}+180-3x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}-3x+180=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+180. To find a and b, set up a system to be solved.

$$a+b=-3 ab=-180=-180$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -180.

$$1,-180 2,-90 3,-60 4,-45 5,-36 6,-30 9,-20 10,-18 12,-15$$

Calculate the sum for each pair.

$$1-180=-179 2-90=-88 3-60=-57 4-45=-41 5-36=-31 6-30=-24 9-20=-11 10-18=-8 12-15=-3$$

The solution is the pair that gives sum -3.

$$a=12 b=-15$$

Rewrite -x^{2}-3x+180 as \left(-x^{2}+12x\right)+\left(-15x+180\right).

$$\left(-x^{2}+12x\right)+\left(-15x+180\right)$$

Factor out x in the first and 15 in the second group.

$$x\left(-x+12\right)+15\left(-x+12\right)$$

Factor out common term -x+12 by using distributive property.

$$\left(-x+12\right)\left(x+15\right)$$

To find equation solutions, solve -x+12=0 and x+15=0.

$$x=12 x=-15$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$-5xx+900=15x$$

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

$$-5x^{2}+900=15x$$

Subtract 15x from both sides.

$$-5x^{2}+900-15x=0$$

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.

$$-5x^{2}-15x+900=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -15 for b, and 900 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\left(-5\right)\times 900}}{2\left(-5\right)}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\left(-5\right)\times 900}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-\left(-15\right)±\sqrt{225+20\times 900}}{2\left(-5\right)}$$

Multiply 20 times 900.

$$x=\frac{-\left(-15\right)±\sqrt{225+18000}}{2\left(-5\right)}$$

Add 225 to 18000.

$$x=\frac{-\left(-15\right)±\sqrt{18225}}{2\left(-5\right)}$$

Take the square root of 18225.

$$x=\frac{-\left(-15\right)±135}{2\left(-5\right)}$$

The opposite of -15 is 15.

$$x=\frac{15±135}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{15±135}{-10}$$

Now solve the equation x=\frac{15±135}{-10} when ± is plus. Add 15 to 135.

$$x=\frac{150}{-10}$$

Divide 150 by -10.

$$x=-15$$

Now solve the equation x=\frac{15±135}{-10} when ± is minus. Subtract 135 from 15.

$$x=-\frac{120}{-10}$$

Divide -120 by -10.

$$x=12$$

The equation is now solved.

$$x=-15 x=12$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$-5xx+900=15x$$

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

$$-5x^{2}+900=15x$$

Subtract 15x from both sides.

$$-5x^{2}+900-15x=0$$

Subtract 900 from both sides. Anything subtracted from zero gives its negation.

$$-5x^{2}-15x=-900$$

Divide both sides by -5.

$$\frac{-5x^{2}-15x}{-5}=-\frac{900}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\left(-\frac{15}{-5}\right)x=-\frac{900}{-5}$$

Divide -15 by -5.

$$x^{2}+3x=-\frac{900}{-5}$$

Divide -900 by -5.

$$x^{2}+3x=180$$

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

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=180+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=180+\frac{9}{4}$$

Add 180 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{729}{4}$$

Factor x^{2}+3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{729}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{729}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{27}{2} x+\frac{3}{2}=-\frac{27}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=12 x=-15$$