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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 positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -6.
-1+6=5 -2+3=1
Calculate the sum for each pair.
a=-2 b=3
The solution is the pair that gives sum 1.
\left(2x^{2}-2x\right)+\left(3x-3\right)
Rewrite 2x^{2}+x-3 as \left(2x^{2}-2x\right)+\left(3x-3\right).
2x\left(x-1\right)+3\left(x-1\right)
Factor out 2x in the first and 3 in the second group.
\left(x-1\right)\left(2x+3\right)
Factor out common term x-1 by using distributive property.
x=1 x=-\frac{3}{2}
To find equation solutions, solve x-1=0 and 2x+3=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{-1±\sqrt{1^{2}-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{-1±\sqrt{1-4\times 2\left(-3\right)}}{2\times 2}
Square 1.
x=\frac{-1±\sqrt{1-8\left(-3\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-1±\sqrt{1+24}}{2\times 2}
Multiply -8 times -3.
x=\frac{-1±\sqrt{25}}{2\times 2}
Add 1 to 24.
x=\frac{-1±5}{2\times 2}
Take the square root of 25.
x=\frac{-1±5}{4}
Multiply 2 times 2.
x=\frac{4}{4}
Now solve the equation x=\frac{-1±5}{4} when ± is plus. Add -1 to 5.
x=1
Divide 4 by 4.
x=-\frac{6}{4}
Now solve the equation x=\frac{-1±5}{4} when ± is minus. Subtract 5 from -1.
x=-\frac{3}{2}
Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.
x=1 x=-\frac{3}{2}
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+\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=1 x=-\frac{3}{2}
Subtract \frac{1}{4} from 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} = -1.500 s = -\frac{1}{4} + \frac{5}{4} = 1
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.