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3x^{2}+72-33x=0
Subtract 33x from both sides.
x^{2}+24-11x=0
Divide both sides by 3.
x^{2}-11x+24=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=-11 ab=1\times 24=24
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+24. To find a and b, set up a system to be solved.
-1,-24 -2,-12 -3,-8 -4,-6
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 24.
-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10
Calculate the sum for each pair.
a=-8 b=-3
The solution is the pair that gives sum -11.
\left(x^{2}-8x\right)+\left(-3x+24\right)
Rewrite x^{2}-11x+24 as \left(x^{2}-8x\right)+\left(-3x+24\right).
x\left(x-8\right)-3\left(x-8\right)
Factor out x in the first and -3 in the second group.
\left(x-8\right)\left(x-3\right)
Factor out common term x-8 by using distributive property.
x=8 x=3
To find equation solutions, solve x-8=0 and x-3=0.
3x^{2}+72-33x=0
Subtract 33x from both sides.
3x^{2}-33x+72=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.
x=\frac{-\left(-33\right)±\sqrt{\left(-33\right)^{2}-4\times 3\times 72}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -33 for b, and 72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-33\right)±\sqrt{1089-4\times 3\times 72}}{2\times 3}
Square -33.
x=\frac{-\left(-33\right)±\sqrt{1089-12\times 72}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-33\right)±\sqrt{1089-864}}{2\times 3}
Multiply -12 times 72.
x=\frac{-\left(-33\right)±\sqrt{225}}{2\times 3}
Add 1089 to -864.
x=\frac{-\left(-33\right)±15}{2\times 3}
Take the square root of 225.
x=\frac{33±15}{2\times 3}
The opposite of -33 is 33.
x=\frac{33±15}{6}
Multiply 2 times 3.
x=\frac{48}{6}
Now solve the equation x=\frac{33±15}{6} when ± is plus. Add 33 to 15.
x=8
Divide 48 by 6.
x=\frac{18}{6}
Now solve the equation x=\frac{33±15}{6} when ± is minus. Subtract 15 from 33.
x=3
Divide 18 by 6.
x=8 x=3
The equation is now solved.
3x^{2}+72-33x=0
Subtract 33x from both sides.
3x^{2}-33x=-72
Subtract 72 from both sides. Anything subtracted from zero gives its negation.
\frac{3x^{2}-33x}{3}=-\frac{72}{3}
Divide both sides by 3.
x^{2}+\left(-\frac{33}{3}\right)x=-\frac{72}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-11x=-\frac{72}{3}
Divide -33 by 3.
x^{2}-11x=-24
Divide -72 by 3.
x^{2}-11x+\left(-\frac{11}{2}\right)^{2}=-24+\left(-\frac{11}{2}\right)^{2}
Divide -11, the coefficient of the x term, by 2 to get -\frac{11}{2}. Then add the square of -\frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-11x+\frac{121}{4}=-24+\frac{121}{4}
Square -\frac{11}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-11x+\frac{121}{4}=\frac{25}{4}
Add -24 to \frac{121}{4}.
\left(x-\frac{11}{2}\right)^{2}=\frac{25}{4}
Factor x^{2}-11x+\frac{121}{4}. 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{11}{2}\right)^{2}}=\sqrt{\frac{25}{4}}
Take the square root of both sides of the equation.
x-\frac{11}{2}=\frac{5}{2} x-\frac{11}{2}=-\frac{5}{2}
Simplify.
x=8 x=3
Add \frac{11}{2} to both sides of the equation.