Solve for x
x=-\frac{1}{3}\approx -0.333333333
x = \frac{3}{2} = 1\frac{1}{2} = 1.5
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a+b=-7 ab=6\left(-3\right)=-18
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6x^{2}+ax+bx-3. 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 negative, the negative number has greater absolute value than the positive. 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=-9 b=2
The solution is the pair that gives sum -7.
\left(6x^{2}-9x\right)+\left(2x-3\right)
Rewrite 6x^{2}-7x-3 as \left(6x^{2}-9x\right)+\left(2x-3\right).
3x\left(2x-3\right)+2x-3
Factor out 3x in 6x^{2}-9x.
\left(2x-3\right)\left(3x+1\right)
Factor out common term 2x-3 by using distributive property.
x=\frac{3}{2} x=-\frac{1}{3}
To find equation solutions, solve 2x-3=0 and 3x+1=0.
6x^{2}-7x-3=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 6\left(-3\right)}}{2\times 6}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -7 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-7\right)±\sqrt{49-4\times 6\left(-3\right)}}{2\times 6}
Square -7.
x=\frac{-\left(-7\right)±\sqrt{49-24\left(-3\right)}}{2\times 6}
Multiply -4 times 6.
x=\frac{-\left(-7\right)±\sqrt{49+72}}{2\times 6}
Multiply -24 times -3.
x=\frac{-\left(-7\right)±\sqrt{121}}{2\times 6}
Add 49 to 72.
x=\frac{-\left(-7\right)±11}{2\times 6}
Take the square root of 121.
x=\frac{7±11}{2\times 6}
The opposite of -7 is 7.
x=\frac{7±11}{12}
Multiply 2 times 6.
x=\frac{18}{12}
Now solve the equation x=\frac{7±11}{12} when ± is plus. Add 7 to 11.
x=\frac{3}{2}
Reduce the fraction \frac{18}{12} to lowest terms by extracting and canceling out 6.
x=-\frac{4}{12}
Now solve the equation x=\frac{7±11}{12} when ± is minus. Subtract 11 from 7.
x=-\frac{1}{3}
Reduce the fraction \frac{-4}{12} to lowest terms by extracting and canceling out 4.
x=\frac{3}{2} x=-\frac{1}{3}
The equation is now solved.
6x^{2}-7x-3=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.
6x^{2}-7x-3-\left(-3\right)=-\left(-3\right)
Add 3 to both sides of the equation.
6x^{2}-7x=-\left(-3\right)
Subtracting -3 from itself leaves 0.
6x^{2}-7x=3
Subtract -3 from 0.
\frac{6x^{2}-7x}{6}=\frac{3}{6}
Divide both sides by 6.
x^{2}-\frac{7}{6}x=\frac{3}{6}
Dividing by 6 undoes the multiplication by 6.
x^{2}-\frac{7}{6}x=\frac{1}{2}
Reduce the fraction \frac{3}{6} to lowest terms by extracting and canceling out 3.
x^{2}-\frac{7}{6}x+\left(-\frac{7}{12}\right)^{2}=\frac{1}{2}+\left(-\frac{7}{12}\right)^{2}
Divide -\frac{7}{6}, the coefficient of the x term, by 2 to get -\frac{7}{12}. Then add the square of -\frac{7}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{7}{6}x+\frac{49}{144}=\frac{1}{2}+\frac{49}{144}
Square -\frac{7}{12} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{7}{6}x+\frac{49}{144}=\frac{121}{144}
Add \frac{1}{2} to \frac{49}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{7}{12}\right)^{2}=\frac{121}{144}
Factor x^{2}-\frac{7}{6}x+\frac{49}{144}. 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}{12}\right)^{2}}=\sqrt{\frac{121}{144}}
Take the square root of both sides of the equation.
x-\frac{7}{12}=\frac{11}{12} x-\frac{7}{12}=-\frac{11}{12}
Simplify.
x=\frac{3}{2} x=-\frac{1}{3}
Add \frac{7}{12} to both sides of the equation.
x ^ 2 -\frac{7}{6}x -\frac{1}{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 6
r + s = \frac{7}{6} rs = -\frac{1}{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}{12} - u s = \frac{7}{12} + u
Two numbers r and s sum up to \frac{7}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{6} = \frac{7}{12}. 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}{12} - u) (\frac{7}{12} + u) = -\frac{1}{2}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{2}
\frac{49}{144} - u^2 = -\frac{1}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{1}{2}-\frac{49}{144} = -\frac{121}{144}
Simplify the expression by subtracting \frac{49}{144} on both sides
u^2 = \frac{121}{144} u = \pm\sqrt{\frac{121}{144}} = \pm \frac{11}{12}
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}{12} - \frac{11}{12} = -0.333 s = \frac{7}{12} + \frac{11}{12} = 1.500
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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