Solve 15x-1/210x^2=-270

$$15 x - \frac { 1 } { 2 } 10 x ^ { 2 } = - 270$$

Solve for x

$$x=-6$$

$$x=9$$

Steps Using Factoring By Grouping

Multiply \frac{1}{2} and 10 to get 5.

$$15x-5x^{2}=-270$$

Add 270 to both sides.

$$15x-5x^{2}+270=0$$

Divide both sides by 5.

$$3x-x^{2}+54=0$$

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

$$-x^{2}+3x+54=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+54. To find a and b, set up a system to be solved.

$$a+b=3 ab=-54=-54$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -54.

$$-1,54 -2,27 -3,18 -6,9$$

Calculate the sum for each pair.

$$-1+54=53 -2+27=25 -3+18=15 -6+9=3$$

The solution is the pair that gives sum 3.

$$a=9 b=-6$$

Rewrite -x^{2}+3x+54 as \left(-x^{2}+9x\right)+\left(-6x+54\right).

$$\left(-x^{2}+9x\right)+\left(-6x+54\right)$$

Factor out -x in the first and -6 in the second group.

$$-x\left(x-9\right)-6\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(-x-6\right)$$

To find equation solutions, solve x-9=0 and -x-6=0.

$$x=9 x=-6$$

Steps Using the Quadratic Formula

Multiply \frac{1}{2} and 10 to get 5.

$$15x-5x^{2}=-270$$

Add 270 to both sides.

$$15x-5x^{2}+270=0$$

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.

$$-5x^{2}+15x+270=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 15 for b, and 270 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-15±\sqrt{15^{2}-4\left(-5\right)\times 270}}{2\left(-5\right)}$$

Square 15.

$$x=\frac{-15±\sqrt{225-4\left(-5\right)\times 270}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-15±\sqrt{225+20\times 270}}{2\left(-5\right)}$$

Multiply 20 times 270.

$$x=\frac{-15±\sqrt{225+5400}}{2\left(-5\right)}$$

Add 225 to 5400.

$$x=\frac{-15±\sqrt{5625}}{2\left(-5\right)}$$

Take the square root of 5625.

$$x=\frac{-15±75}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-15±75}{-10}$$

Now solve the equation x=\frac{-15±75}{-10} when ± is plus. Add -15 to 75.

$$x=\frac{60}{-10}$$

Divide 60 by -10.

$$x=-6$$

Now solve the equation x=\frac{-15±75}{-10} when ± is minus. Subtract 75 from -15.

$$x=-\frac{90}{-10}$$

Divide -90 by -10.

$$x=9$$

The equation is now solved.

$$x=-6 x=9$$

Steps for Completing the Square

Multiply \frac{1}{2} and 10 to get 5.

$$15x-5x^{2}=-270$$

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.

$$-5x^{2}+15x=-270$$

Divide both sides by -5.

$$\frac{-5x^{2}+15x}{-5}=-\frac{270}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{15}{-5}x=-\frac{270}{-5}$$

Divide 15 by -5.

$$x^{2}-3x=-\frac{270}{-5}$$

Divide -270 by -5.

$$x^{2}-3x=54$$

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.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=54+\left(-\frac{3}{2}\right)^{2}$$

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

$$x^{2}-3x+\frac{9}{4}=54+\frac{9}{4}$$

Add 54 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{225}{4}$$

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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{225}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{15}{2} x-\frac{3}{2}=-\frac{15}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=9 x=-6$$