Solve left(3x-5right)^2-left(4x-2right)^2=4(9-10x)

$${ \left(3x-5 \right) }^{ 2 } - { \left(4x-2 \right) }^{ 2 } =4(9-10x)$$

Solve for x

$$x=3$$

$$x=\frac{5}{7}\approx 0.714285714$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-5\right)^{2}.

$$9x^{2}-30x+25-\left(4x-2\right)^{2}=4\left(9-10x\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-2\right)^{2}.

$$9x^{2}-30x+25-\left(16x^{2}-16x+4\right)=4\left(9-10x\right)$$

To find the opposite of 16x^{2}-16x+4, find the opposite of each term.

$$9x^{2}-30x+25-16x^{2}+16x-4=4\left(9-10x\right)$$

Combine 9x^{2} and -16x^{2} to get -7x^{2}.

$$-7x^{2}-30x+25+16x-4=4\left(9-10x\right)$$

Combine -30x and 16x to get -14x.

$$-7x^{2}-14x+25-4=4\left(9-10x\right)$$

Subtract 4 from 25 to get 21.

$$-7x^{2}-14x+21=4\left(9-10x\right)$$

Use the distributive property to multiply 4 by 9-10x.

$$-7x^{2}-14x+21=36-40x$$

Subtract 36 from both sides.

$$-7x^{2}-14x+21-36=-40x$$

Subtract 36 from 21 to get -15.

$$-7x^{2}-14x-15=-40x$$

Add 40x to both sides.

$$-7x^{2}-14x-15+40x=0$$

Combine -14x and 40x to get 26x.

$$-7x^{2}+26x-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -7x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=26 ab=-7\left(-15\right)=105$$

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 105.

$$1,105 3,35 5,21 7,15$$

Calculate the sum for each pair.

$$1+105=106 3+35=38 5+21=26 7+15=22$$

The solution is the pair that gives sum 26.

$$a=21 b=5$$

Rewrite -7x^{2}+26x-15 as \left(-7x^{2}+21x\right)+\left(5x-15\right).

$$\left(-7x^{2}+21x\right)+\left(5x-15\right)$$

Factor out 7x in the first and -5 in the second group.

$$7x\left(-x+3\right)-5\left(-x+3\right)$$

Factor out common term -x+3 by using distributive property.

$$\left(-x+3\right)\left(7x-5\right)$$

To find equation solutions, solve -x+3=0 and 7x-5=0.

$$x=3 x=\frac{5}{7}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-5\right)^{2}.

$$9x^{2}-30x+25-\left(4x-2\right)^{2}=4\left(9-10x\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-2\right)^{2}.

$$9x^{2}-30x+25-\left(16x^{2}-16x+4\right)=4\left(9-10x\right)$$

To find the opposite of 16x^{2}-16x+4, find the opposite of each term.

$$9x^{2}-30x+25-16x^{2}+16x-4=4\left(9-10x\right)$$

Combine 9x^{2} and -16x^{2} to get -7x^{2}.

$$-7x^{2}-30x+25+16x-4=4\left(9-10x\right)$$

Combine -30x and 16x to get -14x.

$$-7x^{2}-14x+25-4=4\left(9-10x\right)$$

Subtract 4 from 25 to get 21.

$$-7x^{2}-14x+21=4\left(9-10x\right)$$

Use the distributive property to multiply 4 by 9-10x.

$$-7x^{2}-14x+21=36-40x$$

Subtract 36 from both sides.

$$-7x^{2}-14x+21-36=-40x$$

Subtract 36 from 21 to get -15.

$$-7x^{2}-14x-15=-40x$$

Add 40x to both sides.

$$-7x^{2}-14x-15+40x=0$$

Combine -14x and 40x to get 26x.

$$-7x^{2}+26x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -7 for a, 26 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-26±\sqrt{26^{2}-4\left(-7\right)\left(-15\right)}}{2\left(-7\right)}$$

Square 26.

$$x=\frac{-26±\sqrt{676-4\left(-7\right)\left(-15\right)}}{2\left(-7\right)}$$

Multiply -4 times -7.

$$x=\frac{-26±\sqrt{676+28\left(-15\right)}}{2\left(-7\right)}$$

Multiply 28 times -15.

$$x=\frac{-26±\sqrt{676-420}}{2\left(-7\right)}$$

Add 676 to -420.

$$x=\frac{-26±\sqrt{256}}{2\left(-7\right)}$$

Take the square root of 256.

$$x=\frac{-26±16}{2\left(-7\right)}$$

Multiply 2 times -7.

$$x=\frac{-26±16}{-14}$$

Now solve the equation x=\frac{-26±16}{-14} when ± is plus. Add -26 to 16.

$$x=-\frac{10}{-14}$$

Reduce the fraction \frac{-10}{-14} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{7}$$

Now solve the equation x=\frac{-26±16}{-14} when ± is minus. Subtract 16 from -26.

$$x=-\frac{42}{-14}$$

Divide -42 by -14.

$$x=3$$

The equation is now solved.

$$x=\frac{5}{7} x=3$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-5\right)^{2}.

$$9x^{2}-30x+25-\left(4x-2\right)^{2}=4\left(9-10x\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-2\right)^{2}.

$$9x^{2}-30x+25-\left(16x^{2}-16x+4\right)=4\left(9-10x\right)$$

To find the opposite of 16x^{2}-16x+4, find the opposite of each term.

$$9x^{2}-30x+25-16x^{2}+16x-4=4\left(9-10x\right)$$

Combine 9x^{2} and -16x^{2} to get -7x^{2}.

$$-7x^{2}-30x+25+16x-4=4\left(9-10x\right)$$

Combine -30x and 16x to get -14x.

$$-7x^{2}-14x+25-4=4\left(9-10x\right)$$

Subtract 4 from 25 to get 21.

$$-7x^{2}-14x+21=4\left(9-10x\right)$$

Use the distributive property to multiply 4 by 9-10x.

$$-7x^{2}-14x+21=36-40x$$

Add 40x to both sides.

$$-7x^{2}-14x+21+40x=36$$

Combine -14x and 40x to get 26x.

$$-7x^{2}+26x+21=36$$

Subtract 21 from both sides.

$$-7x^{2}+26x=36-21$$

Subtract 21 from 36 to get 15.

$$-7x^{2}+26x=15$$

Divide both sides by -7.

$$\frac{-7x^{2}+26x}{-7}=\frac{15}{-7}$$

Dividing by -7 undoes the multiplication by -7.

$$x^{2}+\frac{26}{-7}x=\frac{15}{-7}$$

Divide 26 by -7.

$$x^{2}-\frac{26}{7}x=\frac{15}{-7}$$

Divide 15 by -7.

$$x^{2}-\frac{26}{7}x=-\frac{15}{7}$$

Divide -\frac{26}{7}, the coefficient of the x term, by 2 to get -\frac{13}{7}. Then add the square of -\frac{13}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{26}{7}x+\left(-\frac{13}{7}\right)^{2}=-\frac{15}{7}+\left(-\frac{13}{7}\right)^{2}$$

Square -\frac{13}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{26}{7}x+\frac{169}{49}=-\frac{15}{7}+\frac{169}{49}$$

Add -\frac{15}{7} to \frac{169}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{26}{7}x+\frac{169}{49}=\frac{64}{49}$$

Factor x^{2}-\frac{26}{7}x+\frac{169}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{13}{7}\right)^{2}=\frac{64}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{7}\right)^{2}}=\sqrt{\frac{64}{49}}$$

Simplify.

$$x-\frac{13}{7}=\frac{8}{7} x-\frac{13}{7}=-\frac{8}{7}$$

Add \frac{13}{7} to both sides of the equation.

$$x=3 x=\frac{5}{7}$$