Solve -5x^2+20x+60

$$-5 { x }^{ 2 } +20x+60$$

Factor

$$-5\left(x-6\right)\left(x+2\right)$$

Solution Steps

Factor out 5.

$$5\left(-x^{2}+4x+12\right)$$

Consider -x^{2}+4x+12. Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=4 ab=-12=-12$$

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 -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 4.

$$a=6 b=-2$$

Rewrite -x^{2}+4x+12 as \left(-x^{2}+6x\right)+\left(-2x+12\right).

$$\left(-x^{2}+6x\right)+\left(-2x+12\right)$$

Factor out -x in the first and -2 in the second group.

$$-x\left(x-6\right)-2\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(-x-2\right)$$

Rewrite the complete factored expression.

$$5\left(x-6\right)\left(-x-2\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-5x^{2}+20x+60=0$$

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.

$$x=\frac{-20±\sqrt{20^{2}-4\left(-5\right)\times 60}}{2\left(-5\right)}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\left(-5\right)\times 60}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-20±\sqrt{400+20\times 60}}{2\left(-5\right)}$$

Multiply 20 times 60.

$$x=\frac{-20±\sqrt{400+1200}}{2\left(-5\right)}$$

Add 400 to 1200.

$$x=\frac{-20±\sqrt{1600}}{2\left(-5\right)}$$

Take the square root of 1600.

$$x=\frac{-20±40}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-20±40}{-10}$$

Now solve the equation x=\frac{-20±40}{-10} when ± is plus. Add -20 to 40.

$$x=\frac{20}{-10}$$

Divide 20 by -10.

$$x=-2$$

Now solve the equation x=\frac{-20±40}{-10} when ± is minus. Subtract 40 from -20.

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

Divide -60 by -10.

$$x=6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -2 for x_{1} and 6 for x_{2}.

$$-5x^{2}+20x+60=-5\left(x-\left(-2\right)\right)\left(x-6\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$-5x^{2}+20x+60=-5\left(x+2\right)\left(x-6\right)$$

Evaluate

$$-5\left(x-6\right)\left(x+2\right)$$