Solve (3-2x)^2-(5-x)(5+x)=-20

$$( 3 - 2 x ) ^ { 2 } - ( 5 - x ) ( 5 + x ) = - 20$$

Solve for x

$$x=\frac{2}{5}=0.4$$

$$x=2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-2x\right)^{2}.

$$9-12x+4x^{2}-\left(5-x\right)\left(5+x\right)=-20$$

Consider \left(5-x\right)\left(5+x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 5.

$$9-12x+4x^{2}-\left(25-x^{2}\right)=-20$$

To find the opposite of 25-x^{2}, find the opposite of each term.

$$9-12x+4x^{2}-25+x^{2}=-20$$

Subtract 25 from 9 to get -16.

$$-16-12x+4x^{2}+x^{2}=-20$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$-16-12x+5x^{2}=-20$$

Add 20 to both sides.

$$-16-12x+5x^{2}+20=0$$

Add -16 and 20 to get 4.

$$4-12x+5x^{2}=0$$

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

$$5x^{2}-12x+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-12 ab=5\times 4=20$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. 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 5x^{2}-12x+4 as \left(5x^{2}-10x\right)+\left(-2x+4\right).

$$\left(5x^{2}-10x\right)+\left(-2x+4\right)$$

Factor out 5x in the first and -2 in the second group.

$$5x\left(x-2\right)-2\left(x-2\right)$$

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

$$\left(x-2\right)\left(5x-2\right)$$

To find equation solutions, solve x-2=0 and 5x-2=0.

$$x=2 x=\frac{2}{5}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-2x\right)^{2}.

$$9-12x+4x^{2}-\left(5-x\right)\left(5+x\right)=-20$$

Consider \left(5-x\right)\left(5+x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 5.

$$9-12x+4x^{2}-\left(25-x^{2}\right)=-20$$

To find the opposite of 25-x^{2}, find the opposite of each term.

$$9-12x+4x^{2}-25+x^{2}=-20$$

Subtract 25 from 9 to get -16.

$$-16-12x+4x^{2}+x^{2}=-20$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$-16-12x+5x^{2}=-20$$

Add 20 to both sides.

$$-16-12x+5x^{2}+20=0$$

Add -16 and 20 to get 4.

$$4-12x+5x^{2}=0$$

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.

$$5x^{2}-12x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -12 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 5\times 4}}{2\times 5}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 5\times 4}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-12\right)±\sqrt{144-20\times 4}}{2\times 5}$$

Multiply -20 times 4.

$$x=\frac{-\left(-12\right)±\sqrt{144-80}}{2\times 5}$$

Add 144 to -80.

$$x=\frac{-\left(-12\right)±\sqrt{64}}{2\times 5}$$

Take the square root of 64.

$$x=\frac{-\left(-12\right)±8}{2\times 5}$$

The opposite of -12 is 12.

$$x=\frac{12±8}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{12±8}{10}$$

Now solve the equation x=\frac{12±8}{10} when ± is plus. Add 12 to 8.

$$x=\frac{20}{10}$$

Divide 20 by 10.

$$x=2$$

Now solve the equation x=\frac{12±8}{10} when ± is minus. Subtract 8 from 12.

$$x=\frac{4}{10}$$

Reduce the fraction \frac{4}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{2}{5}$$

The equation is now solved.

$$x=2 x=\frac{2}{5}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-2x\right)^{2}.

$$9-12x+4x^{2}-\left(5-x\right)\left(5+x\right)=-20$$

Consider \left(5-x\right)\left(5+x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 5.

$$9-12x+4x^{2}-\left(25-x^{2}\right)=-20$$

To find the opposite of 25-x^{2}, find the opposite of each term.

$$9-12x+4x^{2}-25+x^{2}=-20$$

Subtract 25 from 9 to get -16.

$$-16-12x+4x^{2}+x^{2}=-20$$

Combine 4x^{2} and x^{2} to get 5x^{2}.

$$-16-12x+5x^{2}=-20$$

Add 16 to both sides.

$$-12x+5x^{2}=-20+16$$

Add -20 and 16 to get -4.

$$-12x+5x^{2}=-4$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$5x^{2}-12x=-4$$

Divide both sides by 5.

$$\frac{5x^{2}-12x}{5}=-\frac{4}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{12}{5}x=-\frac{4}{5}$$

Divide -\frac{12}{5}, the coefficient of the x term, by 2 to get -\frac{6}{5}. Then add the square of -\frac{6}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{12}{5}x+\left(-\frac{6}{5}\right)^{2}=-\frac{4}{5}+\left(-\frac{6}{5}\right)^{2}$$

Square -\frac{6}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{12}{5}x+\frac{36}{25}=-\frac{4}{5}+\frac{36}{25}$$

Add -\frac{4}{5} to \frac{36}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{12}{5}x+\frac{36}{25}=\frac{16}{25}$$

Factor x^{2}-\frac{12}{5}x+\frac{36}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{6}{5}\right)^{2}=\frac{16}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{6}{5}\right)^{2}}=\sqrt{\frac{16}{25}}$$

Simplify.

$$x-\frac{6}{5}=\frac{4}{5} x-\frac{6}{5}=-\frac{4}{5}$$

Add \frac{6}{5} to both sides of the equation.

$$x=2 x=\frac{2}{5}$$