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$2 \exponential{x}{2} + 3 x - 20 = 0 $
Solve for x
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a+b=3 ab=2\left(-20\right)=-40
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-20. To find a and b, set up a system to be solved.
-1,40 -2,20 -4,10 -5,8
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 -40.
-1+40=39 -2+20=18 -4+10=6 -5+8=3
Calculate the sum for each pair.
a=-5 b=8
The solution is the pair that gives sum 3.
\left(2x^{2}-5x\right)+\left(8x-20\right)
Rewrite 2x^{2}+3x-20 as \left(2x^{2}-5x\right)+\left(8x-20\right).
x\left(2x-5\right)+4\left(2x-5\right)
Factor out x in the first and 4 in the second group.
\left(2x-5\right)\left(x+4\right)
Factor out common term 2x-5 by using distributive property.
x=\frac{5}{2} x=-4
To find equation solutions, solve 2x-5=0 and x+4=0.
2x^{2}+3x-20=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{-3±\sqrt{3^{2}-4\times 2\left(-20\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 3 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-3±\sqrt{9-4\times 2\left(-20\right)}}{2\times 2}
Square 3.
x=\frac{-3±\sqrt{9-8\left(-20\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-3±\sqrt{9+160}}{2\times 2}
Multiply -8 times -20.
x=\frac{-3±\sqrt{169}}{2\times 2}
Add 9 to 160.
x=\frac{-3±13}{2\times 2}
Take the square root of 169.
x=\frac{-3±13}{4}
Multiply 2 times 2.
x=\frac{10}{4}
Now solve the equation x=\frac{-3±13}{4} when ± is plus. Add -3 to 13.
x=\frac{5}{2}
Reduce the fraction \frac{10}{4} to lowest terms by extracting and canceling out 2.
x=-\frac{16}{4}
Now solve the equation x=\frac{-3±13}{4} when ± is minus. Subtract 13 from -3.
x=-4
Divide -16 by 4.
x=\frac{5}{2} x=-4
The equation is now solved.
2x^{2}+3x-20=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}+3x-20-\left(-20\right)=-\left(-20\right)
Add 20 to both sides of the equation.
2x^{2}+3x=-\left(-20\right)
Subtracting -20 from itself leaves 0.
2x^{2}+3x=20
Subtract -20 from 0.
\frac{2x^{2}+3x}{2}=\frac{20}{2}
Divide both sides by 2.
x^{2}+\frac{3}{2}x=\frac{20}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}+\frac{3}{2}x=10
Divide 20 by 2.
x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=10+\left(\frac{3}{4}\right)^{2}
Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{3}{2}x+\frac{9}{16}=10+\frac{9}{16}
Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{169}{16}
Add 10 to \frac{9}{16}.
\left(x+\frac{3}{4}\right)^{2}=\frac{169}{16}
Factor x^{2}+\frac{3}{2}x+\frac{9}{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{3}{4}\right)^{2}}=\sqrt{\frac{169}{16}}
Take the square root of both sides of the equation.
x+\frac{3}{4}=\frac{13}{4} x+\frac{3}{4}=-\frac{13}{4}
Simplify.
x=\frac{5}{2} x=-4
Subtract \frac{3}{4} from both sides of the equation.
x ^ 2 +\frac{3}{2}x -10 = 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{3}{2} rs = -10
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{3}{4} - u s = -\frac{3}{4} + u
Two numbers r and s sum up to -\frac{3}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{3}{2} = -\frac{3}{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{3}{4} - u) (-\frac{3}{4} + u) = -10
To solve for unknown quantity u, substitute these in the product equation rs = -10
\frac{9}{16} - u^2 = -10
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -10-\frac{9}{16} = -\frac{169}{16}
Simplify the expression by subtracting \frac{9}{16} on both sides
u^2 = \frac{169}{16} u = \pm\sqrt{\frac{169}{16}} = \pm \frac{13}{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{3}{4} - \frac{13}{4} = -4 s = -\frac{3}{4} + \frac{13}{4} = 2.500
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.