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a+b=-1 ab=2\left(-3\right)=-6
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-3. To find a and b, set up a system to be solved.
1,-6 2,-3
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 -6.
1-6=-5 2-3=-1
Calculate the sum for each pair.
a=-3 b=2
The solution is the pair that gives sum -1.
\left(2x^{2}-3x\right)+\left(2x-3\right)
Rewrite 2x^{2}-x-3 as \left(2x^{2}-3x\right)+\left(2x-3\right).
x\left(2x-3\right)+2x-3
Factor out x in 2x^{2}-3x.
\left(2x-3\right)\left(x+1\right)
Factor out common term 2x-3 by using distributive property.
x=\frac{3}{2} x=-1
To find equation solutions, solve 2x-3=0 and x+1=0.
2x^{2}-x-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{-\left(-1\right)±\sqrt{1-4\times 2\left(-3\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -1 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-1\right)±\sqrt{1-8\left(-3\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-1\right)±\sqrt{1+24}}{2\times 2}
Multiply -8 times -3.
x=\frac{-\left(-1\right)±\sqrt{25}}{2\times 2}
Add 1 to 24.
x=\frac{-\left(-1\right)±5}{2\times 2}
Take the square root of 25.
x=\frac{1±5}{2\times 2}
The opposite of -1 is 1.
x=\frac{1±5}{4}
Multiply 2 times 2.
x=\frac{6}{4}
Now solve the equation x=\frac{1±5}{4} when ± is plus. Add 1 to 5.
x=\frac{3}{2}
Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.
x=\frac{-4}{4}
Now solve the equation x=\frac{1±5}{4} when ± is minus. Subtract 5 from 1.
x=-1
Divide -4 by 4.
x=\frac{3}{2} x=-1
The equation is now solved.
2x^{2}-x-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.
2x^{2}-x-3-\left(-3\right)=-\left(-3\right)
Add 3 to both sides of the equation.
2x^{2}-x=-\left(-3\right)
Subtracting -3 from itself leaves 0.
2x^{2}-x=3
Subtract -3 from 0.
\frac{2x^{2}-x}{2}=\frac{3}{2}
Divide both sides by 2.
x^{2}+\frac{-1}{2}x=\frac{3}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}-\frac{1}{2}x=\frac{3}{2}
Divide -1 by 2.
x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=\frac{3}{2}+\left(-\frac{1}{4}\right)^{2}
Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{3}{2}+\frac{1}{16}
Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{25}{16}
Add \frac{3}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{4}\right)^{2}=\frac{25}{16}
Factor x^{2}-\frac{1}{2}x+\frac{1}{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{1}{4}\right)^{2}}=\sqrt{\frac{25}{16}}
Take the square root of both sides of the equation.
x-\frac{1}{4}=\frac{5}{4} x-\frac{1}{4}=-\frac{5}{4}
Simplify.
x=\frac{3}{2} x=-1
Add \frac{1}{4} to both sides of the equation.
x ^ 2 -\frac{1}{2}x -\frac{3}{2} = 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{1}{2} rs = -\frac{3}{2}
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{1}{4} - u s = \frac{1}{4} + u
Two numbers r and s sum up to \frac{1}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{2} = \frac{1}{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{1}{4} - u) (\frac{1}{4} + u) = -\frac{3}{2}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{2}
\frac{1}{16} - u^2 = -\frac{3}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{3}{2}-\frac{1}{16} = -\frac{25}{16}
Simplify the expression by subtracting \frac{1}{16} on both sides
u^2 = \frac{25}{16} u = \pm\sqrt{\frac{25}{16}} = \pm \frac{5}{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{1}{4} - \frac{5}{4} = -1 s = \frac{1}{4} + \frac{5}{4} = 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.