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3x^{2}-18x+32-5=0
Subtract 5 from both sides.
3x^{2}-18x+27=0
Subtract 5 from 32 to get 27.
x^{2}-6x+9=0
Divide both sides by 3.
a+b=-6 ab=1\times 9=9
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+9. To find a and b, set up a system to be solved.
-1,-9 -3,-3
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 9.
-1-9=-10 -3-3=-6
Calculate the sum for each pair.
a=-3 b=-3
The solution is the pair that gives sum -6.
\left(x^{2}-3x\right)+\left(-3x+9\right)
Rewrite x^{2}-6x+9 as \left(x^{2}-3x\right)+\left(-3x+9\right).
x\left(x-3\right)-3\left(x-3\right)
Factor out x in the first and -3 in the second group.
\left(x-3\right)\left(x-3\right)
Factor out common term x-3 by using distributive property.
\left(x-3\right)^{2}
Rewrite as a binomial square.
x=3
To find equation solution, solve x-3=0.
3x^{2}-18x+32=5
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.
3x^{2}-18x+32-5=5-5
Subtract 5 from both sides of the equation.
3x^{2}-18x+32-5=0
Subtracting 5 from itself leaves 0.
3x^{2}-18x+27=0
Subtract 5 from 32.
x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 3\times 27}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -18 for b, and 27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-18\right)±\sqrt{324-4\times 3\times 27}}{2\times 3}
Square -18.
x=\frac{-\left(-18\right)±\sqrt{324-12\times 27}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-18\right)±\sqrt{324-324}}{2\times 3}
Multiply -12 times 27.
x=\frac{-\left(-18\right)±\sqrt{0}}{2\times 3}
Add 324 to -324.
x=-\frac{-18}{2\times 3}
Take the square root of 0.
x=\frac{18}{2\times 3}
The opposite of -18 is 18.
x=\frac{18}{6}
Multiply 2 times 3.
x=3
Divide 18 by 6.
3x^{2}-18x+32=5
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.
3x^{2}-18x+32-32=5-32
Subtract 32 from both sides of the equation.
3x^{2}-18x=5-32
Subtracting 32 from itself leaves 0.
3x^{2}-18x=-27
Subtract 32 from 5.
\frac{3x^{2}-18x}{3}=-\frac{27}{3}
Divide both sides by 3.
x^{2}+\left(-\frac{18}{3}\right)x=-\frac{27}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-6x=-\frac{27}{3}
Divide -18 by 3.
x^{2}-6x=-9
Divide -27 by 3.
x^{2}-6x+\left(-3\right)^{2}=-9+\left(-3\right)^{2}
Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-6x+9=-9+9
Square -3.
x^{2}-6x+9=0
Add -9 to 9.
\left(x-3\right)^{2}=0
Factor x^{2}-6x+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-3\right)^{2}}=\sqrt{0}
Take the square root of both sides of the equation.
x-3=0 x-3=0
Simplify.
x=3 x=3
Add 3 to both sides of the equation.
x=3
The equation is now solved. Solutions are the same.