Solve for x
x = -\frac{3}{2} = -1\frac{1}{2} = -1.5
x=\frac{1}{5}=0.2
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a+b=13 ab=10\left(-3\right)=-30
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 10x^{2}+ax+bx-3. To find a and b, set up a system to be solved.
-1,30 -2,15 -3,10 -5,6
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 -30.
-1+30=29 -2+15=13 -3+10=7 -5+6=1
Calculate the sum for each pair.
a=-2 b=15
The solution is the pair that gives sum 13.
\left(10x^{2}-2x\right)+\left(15x-3\right)
Rewrite 10x^{2}+13x-3 as \left(10x^{2}-2x\right)+\left(15x-3\right).
2x\left(5x-1\right)+3\left(5x-1\right)
Factor out 2x in the first and 3 in the second group.
\left(5x-1\right)\left(2x+3\right)
Factor out common term 5x-1 by using distributive property.
x=\frac{1}{5} x=-\frac{3}{2}
To find equation solutions, solve 5x-1=0 and 2x+3=0.
10x^{2}+13x-3=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 10\left(-3\right)}}{2\times 10}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, 13 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-13±\sqrt{169-4\times 10\left(-3\right)}}{2\times 10}
Square 13.
x=\frac{-13±\sqrt{169-40\left(-3\right)}}{2\times 10}
Multiply -4 times 10.
x=\frac{-13±\sqrt{169+120}}{2\times 10}
Multiply -40 times -3.
x=\frac{-13±\sqrt{289}}{2\times 10}
Add 169 to 120.
x=\frac{-13±17}{2\times 10}
Take the square root of 289.
x=\frac{-13±17}{20}
Multiply 2 times 10.
x=\frac{4}{20}
Now solve the equation x=\frac{-13±17}{20} when ± is plus. Add -13 to 17.
x=\frac{1}{5}
Reduce the fraction \frac{4}{20} to lowest terms by extracting and canceling out 4.
x=-\frac{30}{20}
Now solve the equation x=\frac{-13±17}{20} when ± is minus. Subtract 17 from -13.
x=-\frac{3}{2}
Reduce the fraction \frac{-30}{20} to lowest terms by extracting and canceling out 10.
x=\frac{1}{5} x=-\frac{3}{2}
The equation is now solved.
10x^{2}+13x-3=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}+13x-3-\left(-3\right)=-\left(-3\right)
Add 3 to both sides of the equation.
10x^{2}+13x=-\left(-3\right)
Subtracting -3 from itself leaves 0.
10x^{2}+13x=3
Subtract -3 from 0.
\frac{10x^{2}+13x}{10}=\frac{3}{10}
Divide both sides by 10.
x^{2}+\frac{13}{10}x=\frac{3}{10}
Dividing by 10 undoes the multiplication by 10.
x^{2}+\frac{13}{10}x+\left(\frac{13}{20}\right)^{2}=\frac{3}{10}+\left(\frac{13}{20}\right)^{2}
Divide \frac{13}{10}, the coefficient of the x term, by 2 to get \frac{13}{20}. Then add the square of \frac{13}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{13}{10}x+\frac{169}{400}=\frac{3}{10}+\frac{169}{400}
Square \frac{13}{20} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{13}{10}x+\frac{169}{400}=\frac{289}{400}
Add \frac{3}{10} to \frac{169}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{13}{20}\right)^{2}=\frac{289}{400}
Factor x^{2}+\frac{13}{10}x+\frac{169}{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{13}{20}\right)^{2}}=\sqrt{\frac{289}{400}}
Take the square root of both sides of the equation.
x+\frac{13}{20}=\frac{17}{20} x+\frac{13}{20}=-\frac{17}{20}
Simplify.
x=\frac{1}{5} x=-\frac{3}{2}
Subtract \frac{13}{20} from both sides of the equation.
x ^ 2 +\frac{13}{10}x -\frac{3}{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 10
r + s = -\frac{13}{10} rs = -\frac{3}{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{13}{20} - u s = -\frac{13}{20} + u
Two numbers r and s sum up to -\frac{13}{10} exactly when the average of the two numbers is \frac{1}{2}*-\frac{13}{10} = -\frac{13}{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{13}{20} - u) (-\frac{13}{20} + u) = -\frac{3}{10}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{10}
\frac{169}{400} - u^2 = -\frac{3}{10}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{3}{10}-\frac{169}{400} = -\frac{289}{400}
Simplify the expression by subtracting \frac{169}{400} on both sides
u^2 = \frac{289}{400} u = \pm\sqrt{\frac{289}{400}} = \pm \frac{17}{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{13}{20} - \frac{17}{20} = -1.500 s = -\frac{13}{20} + \frac{17}{20} = 0.200
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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