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a+b=-7 ab=2\times 6=12
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+6. To find a and b, set up a system to be solved.
-1,-12 -2,-6 -3,-4
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 12.
-1-12=-13 -2-6=-8 -3-4=-7
Calculate the sum for each pair.
a=-4 b=-3
The solution is the pair that gives sum -7.
\left(2x^{2}-4x\right)+\left(-3x+6\right)
Rewrite 2x^{2}-7x+6 as \left(2x^{2}-4x\right)+\left(-3x+6\right).
2x\left(x-2\right)-3\left(x-2\right)
Factor out 2x in the first and -3 in the second group.
\left(x-2\right)\left(2x-3\right)
Factor out common term x-2 by using distributive property.
x=2 x=\frac{3}{2}
To find equation solutions, solve x-2=0 and 2x-3=0.
2x^{2}-7x+6=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(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\times 6}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -7 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-7\right)±\sqrt{49-4\times 2\times 6}}{2\times 2}
Square -7.
x=\frac{-\left(-7\right)±\sqrt{49-8\times 6}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-7\right)±\sqrt{49-48}}{2\times 2}
Multiply -8 times 6.
x=\frac{-\left(-7\right)±\sqrt{1}}{2\times 2}
Add 49 to -48.
x=\frac{-\left(-7\right)±1}{2\times 2}
Take the square root of 1.
x=\frac{7±1}{2\times 2}
The opposite of -7 is 7.
x=\frac{7±1}{4}
Multiply 2 times 2.
x=\frac{8}{4}
Now solve the equation x=\frac{7±1}{4} when ± is plus. Add 7 to 1.
x=2
Divide 8 by 4.
x=\frac{6}{4}
Now solve the equation x=\frac{7±1}{4} when ± is minus. Subtract 1 from 7.
x=\frac{3}{2}
Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.
x=2 x=\frac{3}{2}
The equation is now solved.
2x^{2}-7x+6=0
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.
2x^{2}-7x+6-6=-6
Subtract 6 from both sides of the equation.
2x^{2}-7x=-6
Subtracting 6 from itself leaves 0.
\frac{2x^{2}-7x}{2}=-\frac{6}{2}
Divide both sides by 2.
x^{2}-\frac{7}{2}x=-\frac{6}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}-\frac{7}{2}x=-3
Divide -6 by 2.
x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=-3+\left(-\frac{7}{4}\right)^{2}
Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{7}{2}x+\frac{49}{16}=-3+\frac{49}{16}
Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{1}{16}
Add -3 to \frac{49}{16}.
\left(x-\frac{7}{4}\right)^{2}=\frac{1}{16}
Factor x^{2}-\frac{7}{2}x+\frac{49}{16}. 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{7}{4}\right)^{2}}=\sqrt{\frac{1}{16}}
Take the square root of both sides of the equation.
x-\frac{7}{4}=\frac{1}{4} x-\frac{7}{4}=-\frac{1}{4}
Simplify.
x=2 x=\frac{3}{2}
Add \frac{7}{4} to both sides of the equation.