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4x^{2}+5x-6=0
Divide both sides by 2.
a+b=5 ab=4\left(-6\right)=-24
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-6. To find a and b, set up a system to be solved.
-1,24 -2,12 -3,8 -4,6
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 -24.
-1+24=23 -2+12=10 -3+8=5 -4+6=2
Calculate the sum for each pair.
a=-3 b=8
The solution is the pair that gives sum 5.
\left(4x^{2}-3x\right)+\left(8x-6\right)
Rewrite 4x^{2}+5x-6 as \left(4x^{2}-3x\right)+\left(8x-6\right).
x\left(4x-3\right)+2\left(4x-3\right)
Factor out x in the first and 2 in the second group.
\left(4x-3\right)\left(x+2\right)
Factor out common term 4x-3 by using distributive property.
x=\frac{3}{4} x=-2
To find equation solutions, solve 4x-3=0 and x+2=0.
8x^{2}+10x-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{-10±\sqrt{10^{2}-4\times 8\left(-12\right)}}{2\times 8}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 10 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-10±\sqrt{100-4\times 8\left(-12\right)}}{2\times 8}
Square 10.
x=\frac{-10±\sqrt{100-32\left(-12\right)}}{2\times 8}
Multiply -4 times 8.
x=\frac{-10±\sqrt{100+384}}{2\times 8}
Multiply -32 times -12.
x=\frac{-10±\sqrt{484}}{2\times 8}
Add 100 to 384.
x=\frac{-10±22}{2\times 8}
Take the square root of 484.
x=\frac{-10±22}{16}
Multiply 2 times 8.
x=\frac{12}{16}
Now solve the equation x=\frac{-10±22}{16} when ± is plus. Add -10 to 22.
x=\frac{3}{4}
Reduce the fraction \frac{12}{16} to lowest terms by extracting and canceling out 4.
x=-\frac{32}{16}
Now solve the equation x=\frac{-10±22}{16} when ± is minus. Subtract 22 from -10.
x=-2
Divide -32 by 16.
x=\frac{3}{4} x=-2
The equation is now solved.
8x^{2}+10x-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.
8x^{2}+10x-12-\left(-12\right)=-\left(-12\right)
Add 12 to both sides of the equation.
8x^{2}+10x=-\left(-12\right)
Subtracting -12 from itself leaves 0.
8x^{2}+10x=12
Subtract -12 from 0.
\frac{8x^{2}+10x}{8}=\frac{12}{8}
Divide both sides by 8.
x^{2}+\frac{10}{8}x=\frac{12}{8}
Dividing by 8 undoes the multiplication by 8.
x^{2}+\frac{5}{4}x=\frac{12}{8}
Reduce the fraction \frac{10}{8} to lowest terms by extracting and canceling out 2.
x^{2}+\frac{5}{4}x=\frac{3}{2}
Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.
x^{2}+\frac{5}{4}x+\left(\frac{5}{8}\right)^{2}=\frac{3}{2}+\left(\frac{5}{8}\right)^{2}
Divide \frac{5}{4}, the coefficient of the x term, by 2 to get \frac{5}{8}. Then add the square of \frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{3}{2}+\frac{25}{64}
Square \frac{5}{8} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{121}{64}
Add \frac{3}{2} to \frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{5}{8}\right)^{2}=\frac{121}{64}
Factor x^{2}+\frac{5}{4}x+\frac{25}{64}. 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{5}{8}\right)^{2}}=\sqrt{\frac{121}{64}}
Take the square root of both sides of the equation.
x+\frac{5}{8}=\frac{11}{8} x+\frac{5}{8}=-\frac{11}{8}
Simplify.
x=\frac{3}{4} x=-2
Subtract \frac{5}{8} from both sides of the equation.
x ^ 2 +\frac{5}{4}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 8
r + s = -\frac{5}{4} 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{5}{8} - u s = -\frac{5}{8} + u
Two numbers r and s sum up to -\frac{5}{4} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{4} = -\frac{5}{8}. 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{5}{8} - u) (-\frac{5}{8} + u) = -\frac{3}{2}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{2}
\frac{25}{64} - u^2 = -\frac{3}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{3}{2}-\frac{25}{64} = -\frac{121}{64}
Simplify the expression by subtracting \frac{25}{64} on both sides
u^2 = \frac{121}{64} u = \pm\sqrt{\frac{121}{64}} = \pm \frac{11}{8}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{5}{8} - \frac{11}{8} = -2 s = -\frac{5}{8} + \frac{11}{8} = 0.750
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.