Solve 1500-6x^2-90x=0

$$1500 - 6 x ^ { 2 } - 90 x = 0$$

Solve for x

$$x=-25$$

$$x=10$$

Steps Using Factoring By Grouping

Divide both sides by 6.

$$250-x^{2}-15x=0$$

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

$$-x^{2}-15x+250=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+250. To find a and b, set up a system to be solved.

$$a+b=-15 ab=-250=-250$$

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 -250.

$$1,-250 2,-125 5,-50 10,-25$$

Calculate the sum for each pair.

$$1-250=-249 2-125=-123 5-50=-45 10-25=-15$$

The solution is the pair that gives sum -15.

$$a=10 b=-25$$

Rewrite -x^{2}-15x+250 as \left(-x^{2}+10x\right)+\left(-25x+250\right).

$$\left(-x^{2}+10x\right)+\left(-25x+250\right)$$

Factor out x in the first and 25 in the second group.

$$x\left(-x+10\right)+25\left(-x+10\right)$$

Factor out common term -x+10 by using distributive property.

$$\left(-x+10\right)\left(x+25\right)$$

To find equation solutions, solve -x+10=0 and x+25=0.

$$x=10 x=-25$$

Steps Using the Quadratic Formula

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.

$$-6x^{2}-90x+1500=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, -90 for b, and 1500 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-90\right)±\sqrt{\left(-90\right)^{2}-4\left(-6\right)\times 1500}}{2\left(-6\right)}$$

Square -90.

$$x=\frac{-\left(-90\right)±\sqrt{8100-4\left(-6\right)\times 1500}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-\left(-90\right)±\sqrt{8100+24\times 1500}}{2\left(-6\right)}$$

Multiply 24 times 1500.

$$x=\frac{-\left(-90\right)±\sqrt{8100+36000}}{2\left(-6\right)}$$

Add 8100 to 36000.

$$x=\frac{-\left(-90\right)±\sqrt{44100}}{2\left(-6\right)}$$

Take the square root of 44100.

$$x=\frac{-\left(-90\right)±210}{2\left(-6\right)}$$

The opposite of -90 is 90.

$$x=\frac{90±210}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{90±210}{-12}$$

Now solve the equation x=\frac{90±210}{-12} when ± is plus. Add 90 to 210.

$$x=\frac{300}{-12}$$

Divide 300 by -12.

$$x=-25$$

Now solve the equation x=\frac{90±210}{-12} when ± is minus. Subtract 210 from 90.

$$x=-\frac{120}{-12}$$

Divide -120 by -12.

$$x=10$$

The equation is now solved.

$$x=-25 x=10$$

Steps for Completing the Square

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.

$$-6x^{2}-90x+1500=0$$

Subtract 1500 from both sides of the equation.

$$-6x^{2}-90x+1500-1500=-1500$$

Subtracting 1500 from itself leaves 0.

$$-6x^{2}-90x=-1500$$

Divide both sides by -6.

$$\frac{-6x^{2}-90x}{-6}=-\frac{1500}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\left(-\frac{90}{-6}\right)x=-\frac{1500}{-6}$$

Divide -90 by -6.

$$x^{2}+15x=-\frac{1500}{-6}$$

Divide -1500 by -6.

$$x^{2}+15x=250$$

Divide 15, the coefficient of the x term, by 2 to get \frac{15}{2}. Then add the square of \frac{15}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+15x+\left(\frac{15}{2}\right)^{2}=250+\left(\frac{15}{2}\right)^{2}$$

Square \frac{15}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+15x+\frac{225}{4}=250+\frac{225}{4}$$

Add 250 to \frac{225}{4}.

$$x^{2}+15x+\frac{225}{4}=\frac{1225}{4}$$

Factor x^{2}+15x+\frac{225}{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}.

$$\left(x+\frac{15}{2}\right)^{2}=\frac{1225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{15}{2}\right)^{2}}=\sqrt{\frac{1225}{4}}$$

Simplify.

$$x+\frac{15}{2}=\frac{35}{2} x+\frac{15}{2}=-\frac{35}{2}$$

Subtract \frac{15}{2} from both sides of the equation.

$$x=10 x=-25$$