Solve for x
x=-\frac{2}{5}=-0.4
x = \frac{3}{2} = 1\frac{1}{2} = 1.5
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a+b=-11 ab=10\left(-6\right)=-60
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 10x^{2}+ax+bx-6. 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(10x^{2}-15x\right)+\left(4x-6\right)
Rewrite 10x^{2}-11x-6 as \left(10x^{2}-15x\right)+\left(4x-6\right).
5x\left(2x-3\right)+2\left(2x-3\right)
Factor out 5x in the first and 2 in the second group.
\left(2x-3\right)\left(5x+2\right)
Factor out common term 2x-3 by using distributive property.
x=\frac{3}{2} x=-\frac{2}{5}
To find equation solutions, solve 2x-3=0 and 5x+2=0.
10x^{2}-11x-6=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 10\left(-6\right)}}{2\times 10}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -11 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-11\right)±\sqrt{121-4\times 10\left(-6\right)}}{2\times 10}
Square -11.
x=\frac{-\left(-11\right)±\sqrt{121-40\left(-6\right)}}{2\times 10}
Multiply -4 times 10.
x=\frac{-\left(-11\right)±\sqrt{121+240}}{2\times 10}
Multiply -40 times -6.
x=\frac{-\left(-11\right)±\sqrt{361}}{2\times 10}
Add 121 to 240.
x=\frac{-\left(-11\right)±19}{2\times 10}
Take the square root of 361.
x=\frac{11±19}{2\times 10}
The opposite of -11 is 11.
x=\frac{11±19}{20}
Multiply 2 times 10.
x=\frac{30}{20}
Now solve the equation x=\frac{11±19}{20} when ± is plus. Add 11 to 19.
x=\frac{3}{2}
Reduce the fraction \frac{30}{20} to lowest terms by extracting and canceling out 10.
x=-\frac{8}{20}
Now solve the equation x=\frac{11±19}{20} when ± is minus. Subtract 19 from 11.
x=-\frac{2}{5}
Reduce the fraction \frac{-8}{20} to lowest terms by extracting and canceling out 4.
x=\frac{3}{2} x=-\frac{2}{5}
The equation is now solved.
10x^{2}-11x-6=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.
10x^{2}-11x-6-\left(-6\right)=-\left(-6\right)
Add 6 to both sides of the equation.
10x^{2}-11x=-\left(-6\right)
Subtracting -6 from itself leaves 0.
10x^{2}-11x=6
Subtract -6 from 0.
\frac{10x^{2}-11x}{10}=\frac{6}{10}
Divide both sides by 10.
x^{2}-\frac{11}{10}x=\frac{6}{10}
Dividing by 10 undoes the multiplication by 10.
x^{2}-\frac{11}{10}x=\frac{3}{5}
Reduce the fraction \frac{6}{10} to lowest terms by extracting and canceling out 2.
x^{2}-\frac{11}{10}x+\left(-\frac{11}{20}\right)^{2}=\frac{3}{5}+\left(-\frac{11}{20}\right)^{2}
Divide -\frac{11}{10}, the coefficient of the x term, by 2 to get -\frac{11}{20}. Then add the square of -\frac{11}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{11}{10}x+\frac{121}{400}=\frac{3}{5}+\frac{121}{400}
Square -\frac{11}{20} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{11}{10}x+\frac{121}{400}=\frac{361}{400}
Add \frac{3}{5} to \frac{121}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{11}{20}\right)^{2}=\frac{361}{400}
Factor x^{2}-\frac{11}{10}x+\frac{121}{400}. 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}{20}\right)^{2}}=\sqrt{\frac{361}{400}}
Take the square root of both sides of the equation.
x-\frac{11}{20}=\frac{19}{20} x-\frac{11}{20}=-\frac{19}{20}
Simplify.
x=\frac{3}{2} x=-\frac{2}{5}
Add \frac{11}{20} to both sides of the equation.
x ^ 2 -\frac{11}{10}x -\frac{3}{5} = 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 10
r + s = \frac{11}{10} rs = -\frac{3}{5}
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}{20} - u s = \frac{11}{20} + u
Two numbers r and s sum up to \frac{11}{10} exactly when the average of the two numbers is \frac{1}{2}*\frac{11}{10} = \frac{11}{20}. 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}{20} - u) (\frac{11}{20} + u) = -\frac{3}{5}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{5}
\frac{121}{400} - u^2 = -\frac{3}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{3}{5}-\frac{121}{400} = -\frac{361}{400}
Simplify the expression by subtracting \frac{121}{400} on both sides
u^2 = \frac{361}{400} u = \pm\sqrt{\frac{361}{400}} = \pm \frac{19}{20}
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}{20} - \frac{19}{20} = -0.400 s = \frac{11}{20} + \frac{19}{20} = 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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