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a+b=5 ab=12\left(-2\right)=-24
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 12x^{2}+ax+bx-2. To find a and b, set up a system to be solved.
-1,24 -2,12 -3,8 -4,6
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 -24.
-1+24=23 -2+12=10 -3+8=5 -4+6=2
Calculate the sum for each pair.
a=-3 b=8
The solution is the pair that gives sum 5.
\left(12x^{2}-3x\right)+\left(8x-2\right)
Rewrite 12x^{2}+5x-2 as \left(12x^{2}-3x\right)+\left(8x-2\right).
3x\left(4x-1\right)+2\left(4x-1\right)
Factor out 3x in the first and 2 in the second group.
\left(4x-1\right)\left(3x+2\right)
Factor out common term 4x-1 by using distributive property.
x=\frac{1}{4} x=-\frac{2}{3}
To find equation solutions, solve 4x-1=0 and 3x+2=0.
12x^{2}+5x-2=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{-5±\sqrt{5^{2}-4\times 12\left(-2\right)}}{2\times 12}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, 5 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-5±\sqrt{25-4\times 12\left(-2\right)}}{2\times 12}
Square 5.
x=\frac{-5±\sqrt{25-48\left(-2\right)}}{2\times 12}
Multiply -4 times 12.
x=\frac{-5±\sqrt{25+96}}{2\times 12}
Multiply -48 times -2.
x=\frac{-5±\sqrt{121}}{2\times 12}
Add 25 to 96.
x=\frac{-5±11}{2\times 12}
Take the square root of 121.
x=\frac{-5±11}{24}
Multiply 2 times 12.
x=\frac{6}{24}
Now solve the equation x=\frac{-5±11}{24} when ± is plus. Add -5 to 11.
x=\frac{1}{4}
Reduce the fraction \frac{6}{24} to lowest terms by extracting and canceling out 6.
x=-\frac{16}{24}
Now solve the equation x=\frac{-5±11}{24} when ± is minus. Subtract 11 from -5.
x=-\frac{2}{3}
Reduce the fraction \frac{-16}{24} to lowest terms by extracting and canceling out 8.
x=\frac{1}{4} x=-\frac{2}{3}
The equation is now solved.
12x^{2}+5x-2=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.
12x^{2}+5x-2-\left(-2\right)=-\left(-2\right)
Add 2 to both sides of the equation.
12x^{2}+5x=-\left(-2\right)
Subtracting -2 from itself leaves 0.
12x^{2}+5x=2
Subtract -2 from 0.
\frac{12x^{2}+5x}{12}=\frac{2}{12}
Divide both sides by 12.
x^{2}+\frac{5}{12}x=\frac{2}{12}
Dividing by 12 undoes the multiplication by 12.
x^{2}+\frac{5}{12}x=\frac{1}{6}
Reduce the fraction \frac{2}{12} to lowest terms by extracting and canceling out 2.
x^{2}+\frac{5}{12}x+\left(\frac{5}{24}\right)^{2}=\frac{1}{6}+\left(\frac{5}{24}\right)^{2}
Divide \frac{5}{12}, the coefficient of the x term, by 2 to get \frac{5}{24}. Then add the square of \frac{5}{24} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{5}{12}x+\frac{25}{576}=\frac{1}{6}+\frac{25}{576}
Square \frac{5}{24} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{5}{12}x+\frac{25}{576}=\frac{121}{576}
Add \frac{1}{6} to \frac{25}{576} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{5}{24}\right)^{2}=\frac{121}{576}
Factor x^{2}+\frac{5}{12}x+\frac{25}{576}. 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}{24}\right)^{2}}=\sqrt{\frac{121}{576}}
Take the square root of both sides of the equation.
x+\frac{5}{24}=\frac{11}{24} x+\frac{5}{24}=-\frac{11}{24}
Simplify.
x=\frac{1}{4} x=-\frac{2}{3}
Subtract \frac{5}{24} from both sides of the equation.
x ^ 2 +\frac{5}{12}x -\frac{1}{6} = 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 12
r + s = -\frac{5}{12} rs = -\frac{1}{6}
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{5}{24} - u s = -\frac{5}{24} + u
Two numbers r and s sum up to -\frac{5}{12} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{12} = -\frac{5}{24}. 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{5}{24} - u) (-\frac{5}{24} + u) = -\frac{1}{6}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{6}
\frac{25}{576} - u^2 = -\frac{1}{6}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{1}{6}-\frac{25}{576} = -\frac{121}{576}
Simplify the expression by subtracting \frac{25}{576} on both sides
u^2 = \frac{121}{576} u = \pm\sqrt{\frac{121}{576}} = \pm \frac{11}{24}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{5}{24} - \frac{11}{24} = -0.667 s = -\frac{5}{24} + \frac{11}{24} = 0.250
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.