Solve left(3x-5right)^2-(3x-5)=4+4(9x^2-25)

$${ \left(3x-5 \right) }^{ 2 } -(3x-5)=4+4(9 { x }^{ 2 } -25)$$

Solve for x

$$x = \frac{\sqrt{1633} - 11}{18} \approx 1.633910817$$

$$x=\frac{-\sqrt{1633}-11}{18}\approx -2.856133039$$

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}.

$$9x^{2}-30x+25-\left(3x-5\right)=4+4\left(9x^{2}-25\right)$$

To find the opposite of 3x-5, find the opposite of each term.

$$9x^{2}-30x+25-3x+5=4+4\left(9x^{2}-25\right)$$

Combine -30x and -3x to get -33x.

$$9x^{2}-33x+25+5=4+4\left(9x^{2}-25\right)$$

Add 25 and 5 to get 30.

$$9x^{2}-33x+30=4+4\left(9x^{2}-25\right)$$

Use the distributive property to multiply 4 by 9x^{2}-25.

$$9x^{2}-33x+30=4+36x^{2}-100$$

Subtract 100 from 4 to get -96.

$$9x^{2}-33x+30=-96+36x^{2}$$

Subtract -96 from both sides.

$$9x^{2}-33x+30-\left(-96\right)=36x^{2}$$

The opposite of -96 is 96.

$$9x^{2}-33x+30+96=36x^{2}$$

Subtract 36x^{2} from both sides.

$$9x^{2}-33x+30+96-36x^{2}=0$$

Add 30 and 96 to get 126.

$$9x^{2}-33x+126-36x^{2}=0$$

Combine 9x^{2} and -36x^{2} to get -27x^{2}.

$$-27x^{2}-33x+126=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -27 for a, -33 for b, and 126 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-33\right)±\sqrt{\left(-33\right)^{2}-4\left(-27\right)\times 126}}{2\left(-27\right)}$$

Square -33.

$$x=\frac{-\left(-33\right)±\sqrt{1089-4\left(-27\right)\times 126}}{2\left(-27\right)}$$

Multiply -4 times -27.

$$x=\frac{-\left(-33\right)±\sqrt{1089+108\times 126}}{2\left(-27\right)}$$

Multiply 108 times 126.

$$x=\frac{-\left(-33\right)±\sqrt{1089+13608}}{2\left(-27\right)}$$

Add 1089 to 13608.

$$x=\frac{-\left(-33\right)±\sqrt{14697}}{2\left(-27\right)}$$

Take the square root of 14697.

$$x=\frac{-\left(-33\right)±3\sqrt{1633}}{2\left(-27\right)}$$

The opposite of -33 is 33.

$$x=\frac{33±3\sqrt{1633}}{2\left(-27\right)}$$

Multiply 2 times -27.

$$x=\frac{33±3\sqrt{1633}}{-54}$$

Now solve the equation x=\frac{33±3\sqrt{1633}}{-54} when ± is plus. Add 33 to 3\sqrt{1633}.

$$x=\frac{3\sqrt{1633}+33}{-54}$$

Divide 33+3\sqrt{1633} by -54.

$$x=\frac{-\sqrt{1633}-11}{18}$$

Now solve the equation x=\frac{33±3\sqrt{1633}}{-54} when ± is minus. Subtract 3\sqrt{1633} from 33.

$$x=\frac{33-3\sqrt{1633}}{-54}$$

Divide 33-3\sqrt{1633} by -54.

$$x=\frac{\sqrt{1633}-11}{18}$$

The equation is now solved.

$$x=\frac{-\sqrt{1633}-11}{18} x=\frac{\sqrt{1633}-11}{18}$$

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}.

$$9x^{2}-30x+25-\left(3x-5\right)=4+4\left(9x^{2}-25\right)$$

To find the opposite of 3x-5, find the opposite of each term.

$$9x^{2}-30x+25-3x+5=4+4\left(9x^{2}-25\right)$$

Combine -30x and -3x to get -33x.

$$9x^{2}-33x+25+5=4+4\left(9x^{2}-25\right)$$

Add 25 and 5 to get 30.

$$9x^{2}-33x+30=4+4\left(9x^{2}-25\right)$$

Use the distributive property to multiply 4 by 9x^{2}-25.

$$9x^{2}-33x+30=4+36x^{2}-100$$

Subtract 100 from 4 to get -96.

$$9x^{2}-33x+30=-96+36x^{2}$$

Subtract 36x^{2} from both sides.

$$9x^{2}-33x+30-36x^{2}=-96$$

Combine 9x^{2} and -36x^{2} to get -27x^{2}.

$$-27x^{2}-33x+30=-96$$

Subtract 30 from both sides.

$$-27x^{2}-33x=-96-30$$

Subtract 30 from -96 to get -126.

$$-27x^{2}-33x=-126$$

Divide both sides by -27.

$$\frac{-27x^{2}-33x}{-27}=-\frac{126}{-27}$$

Dividing by -27 undoes the multiplication by -27.

$$x^{2}+\left(-\frac{33}{-27}\right)x=-\frac{126}{-27}$$

Reduce the fraction \frac{-33}{-27} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{11}{9}x=-\frac{126}{-27}$$

Reduce the fraction \frac{-126}{-27} to lowest terms by extracting and canceling out 9.

$$x^{2}+\frac{11}{9}x=\frac{14}{3}$$

Divide \frac{11}{9}, the coefficient of the x term, by 2 to get \frac{11}{18}. Then add the square of \frac{11}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{11}{9}x+\left(\frac{11}{18}\right)^{2}=\frac{14}{3}+\left(\frac{11}{18}\right)^{2}$$

Square \frac{11}{18} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{11}{9}x+\frac{121}{324}=\frac{14}{3}+\frac{121}{324}$$

Add \frac{14}{3} to \frac{121}{324} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{11}{9}x+\frac{121}{324}=\frac{1633}{324}$$

Factor x^{2}+\frac{11}{9}x+\frac{121}{324}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{11}{18}\right)^{2}=\frac{1633}{324}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{18}\right)^{2}}=\sqrt{\frac{1633}{324}}$$

Simplify.

$$x+\frac{11}{18}=\frac{\sqrt{1633}}{18} x+\frac{11}{18}=-\frac{\sqrt{1633}}{18}$$

Subtract \frac{11}{18} from both sides of the equation.

$$x=\frac{\sqrt{1633}-11}{18} x=\frac{-\sqrt{1633}-11}{18}$$