Solve 20(1-x^2)+21x-24=0

$$20(1- { x }^{ 2 } )+21x-24=0$$

Solve for x

$$x=\frac{4}{5}=0.8$$

$$x=\frac{1}{4}=0.25$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 20 by 1-x^{2}.

$$20-20x^{2}+21x-24=0$$

Subtract 24 from 20 to get -4.

$$-4-20x^{2}+21x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-20x^{2}+21x-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -20x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=21 ab=-20\left(-4\right)=80$$

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 80.

$$1,80 2,40 4,20 5,16 8,10$$

Calculate the sum for each pair.

$$1+80=81 2+40=42 4+20=24 5+16=21 8+10=18$$

The solution is the pair that gives sum 21.

$$a=16 b=5$$

Rewrite -20x^{2}+21x-4 as \left(-20x^{2}+16x\right)+\left(5x-4\right).

$$\left(-20x^{2}+16x\right)+\left(5x-4\right)$$

Factor out 4x in the first and -1 in the second group.

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

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

$$\left(-5x+4\right)\left(4x-1\right)$$

To find equation solutions, solve -5x+4=0 and 4x-1=0.

$$x=\frac{4}{5} x=\frac{1}{4}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 20 by 1-x^{2}.

$$20-20x^{2}+21x-24=0$$

Subtract 24 from 20 to get -4.

$$-4-20x^{2}+21x=0$$

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.

$$-20x^{2}+21x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -20 for a, 21 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-21±\sqrt{21^{2}-4\left(-20\right)\left(-4\right)}}{2\left(-20\right)}$$

Square 21.

$$x=\frac{-21±\sqrt{441-4\left(-20\right)\left(-4\right)}}{2\left(-20\right)}$$

Multiply -4 times -20.

$$x=\frac{-21±\sqrt{441+80\left(-4\right)}}{2\left(-20\right)}$$

Multiply 80 times -4.

$$x=\frac{-21±\sqrt{441-320}}{2\left(-20\right)}$$

Add 441 to -320.

$$x=\frac{-21±\sqrt{121}}{2\left(-20\right)}$$

Take the square root of 121.

$$x=\frac{-21±11}{2\left(-20\right)}$$

Multiply 2 times -20.

$$x=\frac{-21±11}{-40}$$

Now solve the equation x=\frac{-21±11}{-40} when ± is plus. Add -21 to 11.

$$x=-\frac{10}{-40}$$

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

$$x=\frac{1}{4}$$

Now solve the equation x=\frac{-21±11}{-40} when ± is minus. Subtract 11 from -21.

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

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

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

The equation is now solved.

$$x=\frac{1}{4} x=\frac{4}{5}$$

Steps for Completing the Square

Use the distributive property to multiply 20 by 1-x^{2}.

$$20-20x^{2}+21x-24=0$$

Subtract 24 from 20 to get -4.

$$-4-20x^{2}+21x=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$-20x^{2}+21x=4$$

Divide both sides by -20.

$$\frac{-20x^{2}+21x}{-20}=\frac{4}{-20}$$

Dividing by -20 undoes the multiplication by -20.

$$x^{2}+\frac{21}{-20}x=\frac{4}{-20}$$

Divide 21 by -20.

$$x^{2}-\frac{21}{20}x=\frac{4}{-20}$$

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

$$x^{2}-\frac{21}{20}x=-\frac{1}{5}$$

Divide -\frac{21}{20}, the coefficient of the x term, by 2 to get -\frac{21}{40}. Then add the square of -\frac{21}{40} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{21}{20}x+\left(-\frac{21}{40}\right)^{2}=-\frac{1}{5}+\left(-\frac{21}{40}\right)^{2}$$

Square -\frac{21}{40} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{21}{20}x+\frac{441}{1600}=-\frac{1}{5}+\frac{441}{1600}$$

Add -\frac{1}{5} to \frac{441}{1600} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{21}{20}x+\frac{441}{1600}=\frac{121}{1600}$$

Factor x^{2}-\frac{21}{20}x+\frac{441}{1600}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{21}{40}\right)^{2}=\frac{121}{1600}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{21}{40}\right)^{2}}=\sqrt{\frac{121}{1600}}$$

Simplify.

$$x-\frac{21}{40}=\frac{11}{40} x-\frac{21}{40}=-\frac{11}{40}$$

Add \frac{21}{40} to both sides of the equation.

$$x=\frac{4}{5} x=\frac{1}{4}$$