Solve for x
x=-\frac{4}{5}=-0.8
x = \frac{3}{2} = 1\frac{1}{2} = 1.5
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a+b=-7 ab=10\left(-12\right)=-120
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 10x^{2}+ax+bx-12. To find a and b, set up a system to be solved.
1,-120 2,-60 3,-40 4,-30 5,-24 6,-20 8,-15 10,-12
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 -120.
1-120=-119 2-60=-58 3-40=-37 4-30=-26 5-24=-19 6-20=-14 8-15=-7 10-12=-2
Calculate the sum for each pair.
a=-15 b=8
The solution is the pair that gives sum -7.
\left(10x^{2}-15x\right)+\left(8x-12\right)
Rewrite 10x^{2}-7x-12 as \left(10x^{2}-15x\right)+\left(8x-12\right).
5x\left(2x-3\right)+4\left(2x-3\right)
Factor out 5x in the first and 4 in the second group.
\left(2x-3\right)\left(5x+4\right)
Factor out common term 2x-3 by using distributive property.
x=\frac{3}{2} x=-\frac{4}{5}
To find equation solutions, solve 2x-3=0 and 5x+4=0.
10x^{2}-7x-12=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(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 10\left(-12\right)}}{2\times 10}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -7 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-7\right)±\sqrt{49-4\times 10\left(-12\right)}}{2\times 10}
Square -7.
x=\frac{-\left(-7\right)±\sqrt{49-40\left(-12\right)}}{2\times 10}
Multiply -4 times 10.
x=\frac{-\left(-7\right)±\sqrt{49+480}}{2\times 10}
Multiply -40 times -12.
x=\frac{-\left(-7\right)±\sqrt{529}}{2\times 10}
Add 49 to 480.
x=\frac{-\left(-7\right)±23}{2\times 10}
Take the square root of 529.
x=\frac{7±23}{2\times 10}
The opposite of -7 is 7.
x=\frac{7±23}{20}
Multiply 2 times 10.
x=\frac{30}{20}
Now solve the equation x=\frac{7±23}{20} when ± is plus. Add 7 to 23.
x=\frac{3}{2}
Reduce the fraction \frac{30}{20} to lowest terms by extracting and canceling out 10.
x=-\frac{16}{20}
Now solve the equation x=\frac{7±23}{20} when ± is minus. Subtract 23 from 7.
x=-\frac{4}{5}
Reduce the fraction \frac{-16}{20} to lowest terms by extracting and canceling out 4.
x=\frac{3}{2} x=-\frac{4}{5}
The equation is now solved.
10x^{2}-7x-12=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}-7x-12-\left(-12\right)=-\left(-12\right)
Add 12 to both sides of the equation.
10x^{2}-7x=-\left(-12\right)
Subtracting -12 from itself leaves 0.
10x^{2}-7x=12
Subtract -12 from 0.
\frac{10x^{2}-7x}{10}=\frac{12}{10}
Divide both sides by 10.
x^{2}-\frac{7}{10}x=\frac{12}{10}
Dividing by 10 undoes the multiplication by 10.
x^{2}-\frac{7}{10}x=\frac{6}{5}
Reduce the fraction \frac{12}{10} to lowest terms by extracting and canceling out 2.
x^{2}-\frac{7}{10}x+\left(-\frac{7}{20}\right)^{2}=\frac{6}{5}+\left(-\frac{7}{20}\right)^{2}
Divide -\frac{7}{10}, the coefficient of the x term, by 2 to get -\frac{7}{20}. Then add the square of -\frac{7}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{7}{10}x+\frac{49}{400}=\frac{6}{5}+\frac{49}{400}
Square -\frac{7}{20} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{7}{10}x+\frac{49}{400}=\frac{529}{400}
Add \frac{6}{5} to \frac{49}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{7}{20}\right)^{2}=\frac{529}{400}
Factor x^{2}-\frac{7}{10}x+\frac{49}{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{7}{20}\right)^{2}}=\sqrt{\frac{529}{400}}
Take the square root of both sides of the equation.
x-\frac{7}{20}=\frac{23}{20} x-\frac{7}{20}=-\frac{23}{20}
Simplify.
x=\frac{3}{2} x=-\frac{4}{5}
Add \frac{7}{20} to both sides of the equation.
x ^ 2 -\frac{7}{10}x -\frac{6}{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{7}{10} rs = -\frac{6}{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{7}{20} - u s = \frac{7}{20} + u
Two numbers r and s sum up to \frac{7}{10} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{10} = \frac{7}{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{7}{20} - u) (\frac{7}{20} + u) = -\frac{6}{5}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{6}{5}
\frac{49}{400} - u^2 = -\frac{6}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{6}{5}-\frac{49}{400} = -\frac{529}{400}
Simplify the expression by subtracting \frac{49}{400} on both sides
u^2 = \frac{529}{400} u = \pm\sqrt{\frac{529}{400}} = \pm \frac{23}{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{7}{20} - \frac{23}{20} = -0.800 s = \frac{7}{20} + \frac{23}{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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