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3x^{2}-15x+12=0
Add 12 to both sides.
x^{2}-5x+4=0
Divide both sides by 3.
a+b=-5 ab=1\times 4=4
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+4. To find a and b, set up a system to be solved.
-1,-4 -2,-2
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 4.
-1-4=-5 -2-2=-4
Calculate the sum for each pair.
a=-4 b=-1
The solution is the pair that gives sum -5.
\left(x^{2}-4x\right)+\left(-x+4\right)
Rewrite x^{2}-5x+4 as \left(x^{2}-4x\right)+\left(-x+4\right).
x\left(x-4\right)-\left(x-4\right)
Factor out x in the first and -1 in the second group.
\left(x-4\right)\left(x-1\right)
Factor out common term x-4 by using distributive property.
x=4 x=1
To find equation solutions, solve x-4=0 and x-1=0.
3x^{2}-15x=-12
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}-15x-\left(-12\right)=-12-\left(-12\right)
Add 12 to both sides of the equation.
3x^{2}-15x-\left(-12\right)=0
Subtracting -12 from itself leaves 0.
3x^{2}-15x+12=0
Subtract -12 from 0.
x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 3\times 12}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -15 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-15\right)±\sqrt{225-4\times 3\times 12}}{2\times 3}
Square -15.
x=\frac{-\left(-15\right)±\sqrt{225-12\times 12}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-15\right)±\sqrt{225-144}}{2\times 3}
Multiply -12 times 12.
x=\frac{-\left(-15\right)±\sqrt{81}}{2\times 3}
Add 225 to -144.
x=\frac{-\left(-15\right)±9}{2\times 3}
Take the square root of 81.
x=\frac{15±9}{2\times 3}
The opposite of -15 is 15.
x=\frac{15±9}{6}
Multiply 2 times 3.
x=\frac{24}{6}
Now solve the equation x=\frac{15±9}{6} when ± is plus. Add 15 to 9.
x=4
Divide 24 by 6.
x=\frac{6}{6}
Now solve the equation x=\frac{15±9}{6} when ± is minus. Subtract 9 from 15.
x=1
Divide 6 by 6.
x=4 x=1
The equation is now solved.
3x^{2}-15x=-12
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.
\frac{3x^{2}-15x}{3}=-\frac{12}{3}
Divide both sides by 3.
x^{2}+\left(-\frac{15}{3}\right)x=-\frac{12}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-5x=-\frac{12}{3}
Divide -15 by 3.
x^{2}-5x=-4
Divide -12 by 3.
x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-4+\left(-\frac{5}{2}\right)^{2}
Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-5x+\frac{25}{4}=-4+\frac{25}{4}
Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-5x+\frac{25}{4}=\frac{9}{4}
Add -4 to \frac{25}{4}.
\left(x-\frac{5}{2}\right)^{2}=\frac{9}{4}
Factor x^{2}-5x+\frac{25}{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{5}{2}\right)^{2}}=\sqrt{\frac{9}{4}}
Take the square root of both sides of the equation.
x-\frac{5}{2}=\frac{3}{2} x-\frac{5}{2}=-\frac{3}{2}
Simplify.
x=4 x=1
Add \frac{5}{2} to both sides of the equation.