Solve -2x^2+10x-12

$$-2 { x }^{ 2 } +10x-12$$

Factor

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

Solution Steps

Factor out 2.

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

Consider -x^{2}+5x-6. Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=5 ab=-\left(-6\right)=6$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 5.

$$a=3 b=2$$

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

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

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

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

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

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

Rewrite the complete factored expression.

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

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.

$$-2x^{2}+10x-12=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-2\right)\left(-12\right)}}{2\left(-2\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-2\right)\left(-12\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-10±\sqrt{100+8\left(-12\right)}}{2\left(-2\right)}$$

Multiply 8 times -12.

$$x=\frac{-10±\sqrt{100-96}}{2\left(-2\right)}$$

Add 100 to -96.

$$x=\frac{-10±\sqrt{4}}{2\left(-2\right)}$$

Take the square root of 4.

$$x=\frac{-10±2}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-10±2}{-4}$$

Now solve the equation x=\frac{-10±2}{-4} when ± is plus. Add -10 to 2.

$$x=-\frac{8}{-4}$$

Divide -8 by -4.

$$x=2$$

Now solve the equation x=\frac{-10±2}{-4} when ± is minus. Subtract 2 from -10.

$$x=-\frac{12}{-4}$$

Divide -12 by -4.

$$x=3$$

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 3 for x_{2}.

$$-2x^{2}+10x-12=-2\left(x-2\right)\left(x-3\right)$$

Evaluate

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