Solve -20x^2+50x-20

$$-20 { x }^{ 2 } +50x-20$$

Factor

$$10\left(2-x\right)\left(2x-1\right)$$

Solution Steps

Factor out 10.

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

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

$$a+b=5 ab=-2\left(-2\right)=4$$

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 4.

$$1,4 2,2$$

Calculate the sum for each pair.

$$1+4=5 2+2=4$$

The solution is the pair that gives sum 5.

$$a=4 b=1$$

Rewrite -2x^{2}+5x-2 as \left(-2x^{2}+4x\right)+\left(x-2\right).

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

Factor out 2x in the first and -1 in the second group.

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

Factor out common term -x+2 by using distributive property.

$$\left(-x+2\right)\left(2x-1\right)$$

Rewrite the complete factored expression.

$$10\left(-x+2\right)\left(2x-1\right)$$

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.

$$-20x^{2}+50x-20=0$$

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.

$$x=\frac{-50±\sqrt{50^{2}-4\left(-20\right)\left(-20\right)}}{2\left(-20\right)}$$

Square 50.

$$x=\frac{-50±\sqrt{2500-4\left(-20\right)\left(-20\right)}}{2\left(-20\right)}$$

Multiply -4 times -20.

$$x=\frac{-50±\sqrt{2500+80\left(-20\right)}}{2\left(-20\right)}$$

Multiply 80 times -20.

$$x=\frac{-50±\sqrt{2500-1600}}{2\left(-20\right)}$$

Add 2500 to -1600.

$$x=\frac{-50±\sqrt{900}}{2\left(-20\right)}$$

Take the square root of 900.

$$x=\frac{-50±30}{2\left(-20\right)}$$

Multiply 2 times -20.

$$x=\frac{-50±30}{-40}$$

Now solve the equation x=\frac{-50±30}{-40} when ± is plus. Add -50 to 30.

$$x=-\frac{20}{-40}$$

Reduce the fraction \frac{-20}{-40} to lowest terms by extracting and canceling out 20.

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{-50±30}{-40} when ± is minus. Subtract 30 from -50.

$$x=-\frac{80}{-40}$$

Divide -80 by -40.

$$x=2$$

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

$$-20x^{2}+50x-20=-20\left(x-\frac{1}{2}\right)\left(x-2\right)$$

Subtract \frac{1}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-20x^{2}+50x-20=-20\times \frac{-2x+1}{-2}\left(x-2\right)$$

Cancel out 2, the greatest common factor in -20 and 2.

$$-20x^{2}+50x-20=10\left(-2x+1\right)\left(x-2\right)$$

Evaluate

$$-20x^{2}+50x-20$$