Solve -6x^2-2x+20

$$-6 { x }^{ 2 } -2x+20$$

Factor

$$2\left(-x-2\right)\left(3x-5\right)$$

Solution Steps

Factor out 2.

$$2\left(-3x^{2}-x+10\right)$$

Consider -3x^{2}-x+10. Factor the expression by grouping. First, the expression needs to be rewritten as -3x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-3\times 10=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Calculate the sum for each pair.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

The solution is the pair that gives sum -1.

$$a=5 b=-6$$

Rewrite -3x^{2}-x+10 as \left(-3x^{2}+5x\right)+\left(-6x+10\right).

$$\left(-3x^{2}+5x\right)+\left(-6x+10\right)$$

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

$$-x\left(3x-5\right)-2\left(3x-5\right)$$

Factor out common term 3x-5 by using distributive property.

$$\left(3x-5\right)\left(-x-2\right)$$

Rewrite the complete factored expression.

$$2\left(3x-5\right)\left(-x-2\right)$$

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.

$$-6x^{2}-2x+20=0$$

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.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-6\right)\times 20}}{2\left(-6\right)}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-6\right)\times 20}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-\left(-2\right)±\sqrt{4+24\times 20}}{2\left(-6\right)}$$

Multiply 24 times 20.

$$x=\frac{-\left(-2\right)±\sqrt{4+480}}{2\left(-6\right)}$$

Add 4 to 480.

$$x=\frac{-\left(-2\right)±\sqrt{484}}{2\left(-6\right)}$$

Take the square root of 484.

$$x=\frac{-\left(-2\right)±22}{2\left(-6\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±22}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{2±22}{-12}$$

Now solve the equation x=\frac{2±22}{-12} when ± is plus. Add 2 to 22.

$$x=\frac{24}{-12}$$

Divide 24 by -12.

$$x=-2$$

Now solve the equation x=\frac{2±22}{-12} when ± is minus. Subtract 22 from 2.

$$x=-\frac{20}{-12}$$

Reduce the fraction \frac{-20}{-12} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{3}$$

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 \frac{5}{3} for x_{2}.

$$-6x^{2}-2x+20=-6\left(x-\left(-2\right)\right)\left(x-\frac{5}{3}\right)$$

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

$$-6x^{2}-2x+20=-6\left(x+2\right)\left(x-\frac{5}{3}\right)$$

Subtract \frac{5}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-6x^{2}-2x+20=-6\left(x+2\right)\times \frac{-3x+5}{-3}$$

Cancel out 3, the greatest common factor in -6 and 3.

$$-6x^{2}-2x+20=2\left(x+2\right)\left(-3x+5\right)$$

Evaluate

$$20-2x-6x^{2}$$