Solve -x^2+12x-20

$$- { x }^{ 2 } +12x-20$$

Factor

$$\left(2-x\right)\left(x-10\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=12 ab=-\left(-20\right)=20$$

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 20.

$$1,20 2,10 4,5$$

Calculate the sum for each pair.

$$1+20=21 2+10=12 4+5=9$$

The solution is the pair that gives sum 12.

$$a=10 b=2$$

Rewrite -x^{2}+12x-20 as \left(-x^{2}+10x\right)+\left(2x-20\right).

$$\left(-x^{2}+10x\right)+\left(2x-20\right)$$

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

$$-x\left(x-10\right)+2\left(x-10\right)$$

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

$$\left(x-10\right)\left(-x+2\right)$$

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.

$$-x^{2}+12x-20=0$$

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.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-1\right)\left(-20\right)}}{2\left(-1\right)}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-1\right)\left(-20\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-12±\sqrt{144+4\left(-20\right)}}{2\left(-1\right)}$$

Multiply 4 times -20.

$$x=\frac{-12±\sqrt{144-80}}{2\left(-1\right)}$$

Add 144 to -80.

$$x=\frac{-12±\sqrt{64}}{2\left(-1\right)}$$

Take the square root of 64.

$$x=\frac{-12±8}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-12±8}{-2}$$

Now solve the equation x=\frac{-12±8}{-2} when ± is plus. Add -12 to 8.

$$x=-\frac{4}{-2}$$

Divide -4 by -2.

$$x=2$$

Now solve the equation x=\frac{-12±8}{-2} when ± is minus. Subtract 8 from -12.

$$x=-\frac{20}{-2}$$

Divide -20 by -2.

$$x=10$$

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 10 for x_{2}.

$$-x^{2}+12x-20=-\left(x-2\right)\left(x-10\right)$$

Evaluate

$$\left(2-x\right)\left(x-10\right)$$