Solve -20x^2-9x+20=

$$-20 { x }^{ 2 } -9x+20=$$

Factor

$$\left(4-5x\right)\left(4x+5\right)$$

Solution Steps

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

$$a+b=-9 ab=-20\times 20=-400$$

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 -400.

$$1,-400 2,-200 4,-100 5,-80 8,-50 10,-40 16,-25 20,-20$$

Calculate the sum for each pair.

$$1-400=-399 2-200=-198 4-100=-96 5-80=-75 8-50=-42 10-40=-30 16-25=-9 20-20=0$$

The solution is the pair that gives sum -9.

$$a=16 b=-25$$

Rewrite -20x^{2}-9x+20 as \left(-20x^{2}+16x\right)+\left(-25x+20\right).

$$\left(-20x^{2}+16x\right)+\left(-25x+20\right)$$

Factor out 4x in the first and 5 in the second group.

$$4x\left(-5x+4\right)+5\left(-5x+4\right)$$

Factor out common term -5x+4 by using distributive property.

$$\left(-5x+4\right)\left(4x+5\right)$$

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.

$$-20x^{2}-9x+20=0$$

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.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\left(-20\right)\times 20}}{2\left(-20\right)}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\left(-20\right)\times 20}}{2\left(-20\right)}$$

Multiply -4 times -20.

$$x=\frac{-\left(-9\right)±\sqrt{81+80\times 20}}{2\left(-20\right)}$$

Multiply 80 times 20.

$$x=\frac{-\left(-9\right)±\sqrt{81+1600}}{2\left(-20\right)}$$

Add 81 to 1600.

$$x=\frac{-\left(-9\right)±\sqrt{1681}}{2\left(-20\right)}$$

Take the square root of 1681.

$$x=\frac{-\left(-9\right)±41}{2\left(-20\right)}$$

The opposite of -9 is 9.

$$x=\frac{9±41}{2\left(-20\right)}$$

Multiply 2 times -20.

$$x=\frac{9±41}{-40}$$

Now solve the equation x=\frac{9±41}{-40} when ± is plus. Add 9 to 41.

$$x=\frac{50}{-40}$$

Reduce the fraction \frac{50}{-40} to lowest terms by extracting and canceling out 10.

$$x=-\frac{5}{4}$$

Now solve the equation x=\frac{9±41}{-40} when ± is minus. Subtract 41 from 9.

$$x=-\frac{32}{-40}$$

Reduce the fraction \frac{-32}{-40} to lowest terms by extracting and canceling out 8.

$$x=\frac{4}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{5}{4} for x_{1} and \frac{4}{5} for x_{2}.

$$-20x^{2}-9x+20=-20\left(x-\left(-\frac{5}{4}\right)\right)\left(x-\frac{4}{5}\right)$$

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

$$-20x^{2}-9x+20=-20\left(x+\frac{5}{4}\right)\left(x-\frac{4}{5}\right)$$

Add \frac{5}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-20x^{2}-9x+20=-20\times \frac{-4x-5}{-4}\left(x-\frac{4}{5}\right)$$

Subtract \frac{4}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-20x^{2}-9x+20=-20\times \frac{-4x-5}{-4}\times \frac{-5x+4}{-5}$$

Multiply \frac{-4x-5}{-4} times \frac{-5x+4}{-5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-20x^{2}-9x+20=-20\times \frac{\left(-4x-5\right)\left(-5x+4\right)}{-4\left(-5\right)}$$

Multiply -4 times -5.

$$-20x^{2}-9x+20=-20\times \frac{\left(-4x-5\right)\left(-5x+4\right)}{20}$$

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

$$-20x^{2}-9x+20=-\left(-4x-5\right)\left(-5x+4\right)$$

Evaluate

$$\left(4-5x\right)\left(4x+5\right)$$