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a+b=7 ab=3\left(-6\right)=-18
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-6. To find a and b, set up a system to be solved.
-1,18 -2,9 -3,6
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -18.
-1+18=17 -2+9=7 -3+6=3
Calculate the sum for each pair.
a=-2 b=9
The solution is the pair that gives sum 7.
\left(3x^{2}-2x\right)+\left(9x-6\right)
Rewrite 3x^{2}+7x-6 as \left(3x^{2}-2x\right)+\left(9x-6\right).
x\left(3x-2\right)+3\left(3x-2\right)
Factor out x in the first and 3 in the second group.
\left(3x-2\right)\left(x+3\right)
Factor out common term 3x-2 by using distributive property.
x=\frac{2}{3} x=-3
To find equation solutions, solve 3x-2=0 and x+3=0.
3x^{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{-7±\sqrt{7^{2}-4\times 3\left(-6\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 7 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-7±\sqrt{49-4\times 3\left(-6\right)}}{2\times 3}
Square 7.
x=\frac{-7±\sqrt{49-12\left(-6\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-7±\sqrt{49+72}}{2\times 3}
Multiply -12 times -6.
x=\frac{-7±\sqrt{121}}{2\times 3}
Add 49 to 72.
x=\frac{-7±11}{2\times 3}
Take the square root of 121.
x=\frac{-7±11}{6}
Multiply 2 times 3.
x=\frac{4}{6}
Now solve the equation x=\frac{-7±11}{6} when ± is plus. Add -7 to 11.
x=\frac{2}{3}
Reduce the fraction \frac{4}{6} to lowest terms by extracting and canceling out 2.
x=-\frac{18}{6}
Now solve the equation x=\frac{-7±11}{6} when ± is minus. Subtract 11 from -7.
x=-3
Divide -18 by 6.
x=\frac{2}{3} x=-3
The equation is now solved.
3x^{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.
3x^{2}+7x-6-\left(-6\right)=-\left(-6\right)
Add 6 to both sides of the equation.
3x^{2}+7x=-\left(-6\right)
Subtracting -6 from itself leaves 0.
3x^{2}+7x=6
Subtract -6 from 0.
\frac{3x^{2}+7x}{3}=\frac{6}{3}
Divide both sides by 3.
x^{2}+\frac{7}{3}x=\frac{6}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}+\frac{7}{3}x=2
Divide 6 by 3.
x^{2}+\frac{7}{3}x+\left(\frac{7}{6}\right)^{2}=2+\left(\frac{7}{6}\right)^{2}
Divide \frac{7}{3}, the coefficient of the x term, by 2 to get \frac{7}{6}. Then add the square of \frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{7}{3}x+\frac{49}{36}=2+\frac{49}{36}
Square \frac{7}{6} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{7}{3}x+\frac{49}{36}=\frac{121}{36}
Add 2 to \frac{49}{36}.
\left(x+\frac{7}{6}\right)^{2}=\frac{121}{36}
Factor x^{2}+\frac{7}{3}x+\frac{49}{36}. 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}{6}\right)^{2}}=\sqrt{\frac{121}{36}}
Take the square root of both sides of the equation.
x+\frac{7}{6}=\frac{11}{6} x+\frac{7}{6}=-\frac{11}{6}
Simplify.
x=\frac{2}{3} x=-3
Subtract \frac{7}{6} from both sides of the equation.
x ^ 2 +\frac{7}{3}x -2 = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 3
r + s = -\frac{7}{3} rs = -2
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = -\frac{7}{6} - u s = -\frac{7}{6} + u
Two numbers r and s sum up to -\frac{7}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{3} = -\frac{7}{6}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(-\frac{7}{6} - u) (-\frac{7}{6} + u) = -2
To solve for unknown quantity u, substitute these in the product equation rs = -2
\frac{49}{36} - u^2 = -2
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -2-\frac{49}{36} = -\frac{121}{36}
Simplify the expression by subtracting \frac{49}{36} on both sides
u^2 = \frac{121}{36} u = \pm\sqrt{\frac{121}{36}} = \pm \frac{11}{6}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{7}{6} - \frac{11}{6} = -3 s = -\frac{7}{6} + \frac{11}{6} = 0.667
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.