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a+b=13 ab=2\left(-24\right)=-48
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-24. To find a and b, set up a system to be solved.
-1,48 -2,24 -3,16 -4,12 -6,8
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 -48.
-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2
Calculate the sum for each pair.
a=-3 b=16
The solution is the pair that gives sum 13.
\left(2x^{2}-3x\right)+\left(16x-24\right)
Rewrite 2x^{2}+13x-24 as \left(2x^{2}-3x\right)+\left(16x-24\right).
x\left(2x-3\right)+8\left(2x-3\right)
Factor out x in the first and 8 in the second group.
\left(2x-3\right)\left(x+8\right)
Factor out common term 2x-3 by using distributive property.
x=\frac{3}{2} x=-8
To find equation solutions, solve 2x-3=0 and x+8=0.
2x^{2}+13x-24=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{-13±\sqrt{13^{2}-4\times 2\left(-24\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 13 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-13±\sqrt{169-4\times 2\left(-24\right)}}{2\times 2}
Square 13.
x=\frac{-13±\sqrt{169-8\left(-24\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-13±\sqrt{169+192}}{2\times 2}
Multiply -8 times -24.
x=\frac{-13±\sqrt{361}}{2\times 2}
Add 169 to 192.
x=\frac{-13±19}{2\times 2}
Take the square root of 361.
x=\frac{-13±19}{4}
Multiply 2 times 2.
x=\frac{6}{4}
Now solve the equation x=\frac{-13±19}{4} when ± is plus. Add -13 to 19.
x=\frac{3}{2}
Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.
x=-\frac{32}{4}
Now solve the equation x=\frac{-13±19}{4} when ± is minus. Subtract 19 from -13.
x=-8
Divide -32 by 4.
x=\frac{3}{2} x=-8
The equation is now solved.
2x^{2}+13x-24=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}+13x-24-\left(-24\right)=-\left(-24\right)
Add 24 to both sides of the equation.
2x^{2}+13x=-\left(-24\right)
Subtracting -24 from itself leaves 0.
2x^{2}+13x=24
Subtract -24 from 0.
\frac{2x^{2}+13x}{2}=\frac{24}{2}
Divide both sides by 2.
x^{2}+\frac{13}{2}x=\frac{24}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}+\frac{13}{2}x=12
Divide 24 by 2.
x^{2}+\frac{13}{2}x+\left(\frac{13}{4}\right)^{2}=12+\left(\frac{13}{4}\right)^{2}
Divide \frac{13}{2}, the coefficient of the x term, by 2 to get \frac{13}{4}. Then add the square of \frac{13}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{13}{2}x+\frac{169}{16}=12+\frac{169}{16}
Square \frac{13}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{13}{2}x+\frac{169}{16}=\frac{361}{16}
Add 12 to \frac{169}{16}.
\left(x+\frac{13}{4}\right)^{2}=\frac{361}{16}
Factor x^{2}+\frac{13}{2}x+\frac{169}{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{13}{4}\right)^{2}}=\sqrt{\frac{361}{16}}
Take the square root of both sides of the equation.
x+\frac{13}{4}=\frac{19}{4} x+\frac{13}{4}=-\frac{19}{4}
Simplify.
x=\frac{3}{2} x=-8
Subtract \frac{13}{4} from both sides of the equation.
x ^ 2 +\frac{13}{2}x -12 = 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 2
r + s = -\frac{13}{2} rs = -12
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{13}{4} - u s = -\frac{13}{4} + u
Two numbers r and s sum up to -\frac{13}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{13}{2} = -\frac{13}{4}. 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{13}{4} - u) (-\frac{13}{4} + u) = -12
To solve for unknown quantity u, substitute these in the product equation rs = -12
\frac{169}{16} - u^2 = -12
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -12-\frac{169}{16} = -\frac{361}{16}
Simplify the expression by subtracting \frac{169}{16} on both sides
u^2 = \frac{361}{16} u = \pm\sqrt{\frac{361}{16}} = \pm \frac{19}{4}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{13}{4} - \frac{19}{4} = -8 s = -\frac{13}{4} + \frac{19}{4} = 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.