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a+b=-11 ab=3\left(-20\right)=-60
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-20. To find a and b, set up a system to be solved.
1,-60 2,-30 3,-20 4,-15 5,-12 6,-10
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 -60.
1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4
Calculate the sum for each pair.
a=-15 b=4
The solution is the pair that gives sum -11.
\left(3x^{2}-15x\right)+\left(4x-20\right)
Rewrite 3x^{2}-11x-20 as \left(3x^{2}-15x\right)+\left(4x-20\right).
3x\left(x-5\right)+4\left(x-5\right)
Factor out 3x in the first and 4 in the second group.
\left(x-5\right)\left(3x+4\right)
Factor out common term x-5 by using distributive property.
x=5 x=-\frac{4}{3}
To find equation solutions, solve x-5=0 and 3x+4=0.
3x^{2}-11x-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{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 3\left(-20\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -11 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-11\right)±\sqrt{121-4\times 3\left(-20\right)}}{2\times 3}
Square -11.
x=\frac{-\left(-11\right)±\sqrt{121-12\left(-20\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-11\right)±\sqrt{121+240}}{2\times 3}
Multiply -12 times -20.
x=\frac{-\left(-11\right)±\sqrt{361}}{2\times 3}
Add 121 to 240.
x=\frac{-\left(-11\right)±19}{2\times 3}
Take the square root of 361.
x=\frac{11±19}{2\times 3}
The opposite of -11 is 11.
x=\frac{11±19}{6}
Multiply 2 times 3.
x=\frac{30}{6}
Now solve the equation x=\frac{11±19}{6} when ± is plus. Add 11 to 19.
x=5
Divide 30 by 6.
x=-\frac{8}{6}
Now solve the equation x=\frac{11±19}{6} when ± is minus. Subtract 19 from 11.
x=-\frac{4}{3}
Reduce the fraction \frac{-8}{6} to lowest terms by extracting and canceling out 2.
x=5 x=-\frac{4}{3}
The equation is now solved.
3x^{2}-11x-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.
3x^{2}-11x-20-\left(-20\right)=-\left(-20\right)
Add 20 to both sides of the equation.
3x^{2}-11x=-\left(-20\right)
Subtracting -20 from itself leaves 0.
3x^{2}-11x=20
Subtract -20 from 0.
\frac{3x^{2}-11x}{3}=\frac{20}{3}
Divide both sides by 3.
x^{2}-\frac{11}{3}x=\frac{20}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-\frac{11}{3}x+\left(-\frac{11}{6}\right)^{2}=\frac{20}{3}+\left(-\frac{11}{6}\right)^{2}
Divide -\frac{11}{3}, the coefficient of the x term, by 2 to get -\frac{11}{6}. Then add the square of -\frac{11}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{11}{3}x+\frac{121}{36}=\frac{20}{3}+\frac{121}{36}
Square -\frac{11}{6} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{11}{3}x+\frac{121}{36}=\frac{361}{36}
Add \frac{20}{3} to \frac{121}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{11}{6}\right)^{2}=\frac{361}{36}
Factor x^{2}-\frac{11}{3}x+\frac{121}{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{11}{6}\right)^{2}}=\sqrt{\frac{361}{36}}
Take the square root of both sides of the equation.
x-\frac{11}{6}=\frac{19}{6} x-\frac{11}{6}=-\frac{19}{6}
Simplify.
x=5 x=-\frac{4}{3}
Add \frac{11}{6} to both sides of the equation.
x ^ 2 -\frac{11}{3}x -\frac{20}{3} = 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{11}{3} rs = -\frac{20}{3}
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{11}{6} - u s = \frac{11}{6} + u
Two numbers r and s sum up to \frac{11}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{11}{3} = \frac{11}{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{11}{6} - u) (\frac{11}{6} + u) = -\frac{20}{3}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{20}{3}
\frac{121}{36} - u^2 = -\frac{20}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{20}{3}-\frac{121}{36} = -\frac{361}{36}
Simplify the expression by subtracting \frac{121}{36} on both sides
u^2 = \frac{361}{36} u = \pm\sqrt{\frac{361}{36}} = \pm \frac{19}{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{11}{6} - \frac{19}{6} = -1.333 s = \frac{11}{6} + \frac{19}{6} = 5
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.