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9x^{2}-12x+4-9=0
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-2\right)^{2}.
9x^{2}-12x-5=0
Subtract 9 from 4 to get -5.
a+b=-12 ab=9\left(-5\right)=-45
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx-5. To find a and b, set up a system to be solved.
1,-45 3,-15 5,-9
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 -45.
1-45=-44 3-15=-12 5-9=-4
Calculate the sum for each pair.
a=-15 b=3
The solution is the pair that gives sum -12.
\left(9x^{2}-15x\right)+\left(3x-5\right)
Rewrite 9x^{2}-12x-5 as \left(9x^{2}-15x\right)+\left(3x-5\right).
3x\left(3x-5\right)+3x-5
Factor out 3x in 9x^{2}-15x.
\left(3x-5\right)\left(3x+1\right)
Factor out common term 3x-5 by using distributive property.
x=\frac{5}{3} x=-\frac{1}{3}
To find equation solutions, solve 3x-5=0 and 3x+1=0.
9x^{2}-12x+4-9=0
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-2\right)^{2}.
9x^{2}-12x-5=0
Subtract 9 from 4 to get -5.
x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 9\left(-5\right)}}{2\times 9}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -12 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-12\right)±\sqrt{144-4\times 9\left(-5\right)}}{2\times 9}
Square -12.
x=\frac{-\left(-12\right)±\sqrt{144-36\left(-5\right)}}{2\times 9}
Multiply -4 times 9.
x=\frac{-\left(-12\right)±\sqrt{144+180}}{2\times 9}
Multiply -36 times -5.
x=\frac{-\left(-12\right)±\sqrt{324}}{2\times 9}
Add 144 to 180.
x=\frac{-\left(-12\right)±18}{2\times 9}
Take the square root of 324.
x=\frac{12±18}{2\times 9}
The opposite of -12 is 12.
x=\frac{12±18}{18}
Multiply 2 times 9.
x=\frac{30}{18}
Now solve the equation x=\frac{12±18}{18} when ± is plus. Add 12 to 18.
x=\frac{5}{3}
Reduce the fraction \frac{30}{18} to lowest terms by extracting and canceling out 6.
x=-\frac{6}{18}
Now solve the equation x=\frac{12±18}{18} when ± is minus. Subtract 18 from 12.
x=-\frac{1}{3}
Reduce the fraction \frac{-6}{18} to lowest terms by extracting and canceling out 6.
x=\frac{5}{3} x=-\frac{1}{3}
The equation is now solved.
9x^{2}-12x+4-9=0
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-2\right)^{2}.
9x^{2}-12x-5=0
Subtract 9 from 4 to get -5.
9x^{2}-12x=5
Add 5 to both sides. Anything plus zero gives itself.
\frac{9x^{2}-12x}{9}=\frac{5}{9}
Divide both sides by 9.
x^{2}+\left(-\frac{12}{9}\right)x=\frac{5}{9}
Dividing by 9 undoes the multiplication by 9.
x^{2}-\frac{4}{3}x=\frac{5}{9}
Reduce the fraction \frac{-12}{9} to lowest terms by extracting and canceling out 3.
x^{2}-\frac{4}{3}x+\left(-\frac{2}{3}\right)^{2}=\frac{5}{9}+\left(-\frac{2}{3}\right)^{2}
Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{5+4}{9}
Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{4}{3}x+\frac{4}{9}=1
Add \frac{5}{9} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{2}{3}\right)^{2}=1
Factor x^{2}-\frac{4}{3}x+\frac{4}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{1}
Take the square root of both sides of the equation.
x-\frac{2}{3}=1 x-\frac{2}{3}=-1
Simplify.
x=\frac{5}{3} x=-\frac{1}{3}
Add \frac{2}{3} to both sides of the equation.