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a+b=-12 ab=5\left(-9\right)=-45
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-9. To find a and b, set up a system to be solved.
1,-45 3,-15 5,-9
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 -45.
1-45=-44 3-15=-12 5-9=-4
Calculate the sum for each pair.
a=-15 b=3
The solution is the pair that gives sum -12.
\left(5x^{2}-15x\right)+\left(3x-9\right)
Rewrite 5x^{2}-12x-9 as \left(5x^{2}-15x\right)+\left(3x-9\right).
5x\left(x-3\right)+3\left(x-3\right)
Factor out 5x in the first and 3 in the second group.
\left(x-3\right)\left(5x+3\right)
Factor out common term x-3 by using distributive property.
x=3 x=-\frac{3}{5}
To find equation solutions, solve x-3=0 and 5x+3=0.
5x^{2}-12x-9=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(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 5\left(-9\right)}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -12 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-12\right)±\sqrt{144-4\times 5\left(-9\right)}}{2\times 5}
Square -12.
x=\frac{-\left(-12\right)±\sqrt{144-20\left(-9\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-12\right)±\sqrt{144+180}}{2\times 5}
Multiply -20 times -9.
x=\frac{-\left(-12\right)±\sqrt{324}}{2\times 5}
Add 144 to 180.
x=\frac{-\left(-12\right)±18}{2\times 5}
Take the square root of 324.
x=\frac{12±18}{2\times 5}
The opposite of -12 is 12.
x=\frac{12±18}{10}
Multiply 2 times 5.
x=\frac{30}{10}
Now solve the equation x=\frac{12±18}{10} when ± is plus. Add 12 to 18.
x=3
Divide 30 by 10.
x=-\frac{6}{10}
Now solve the equation x=\frac{12±18}{10} when ± is minus. Subtract 18 from 12.
x=-\frac{3}{5}
Reduce the fraction \frac{-6}{10} to lowest terms by extracting and canceling out 2.
x=3 x=-\frac{3}{5}
The equation is now solved.
5x^{2}-12x-9=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.
5x^{2}-12x-9-\left(-9\right)=-\left(-9\right)
Add 9 to both sides of the equation.
5x^{2}-12x=-\left(-9\right)
Subtracting -9 from itself leaves 0.
5x^{2}-12x=9
Subtract -9 from 0.
\frac{5x^{2}-12x}{5}=\frac{9}{5}
Divide both sides by 5.
x^{2}-\frac{12}{5}x=\frac{9}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-\frac{12}{5}x+\left(-\frac{6}{5}\right)^{2}=\frac{9}{5}+\left(-\frac{6}{5}\right)^{2}
Divide -\frac{12}{5}, the coefficient of the x term, by 2 to get -\frac{6}{5}. Then add the square of -\frac{6}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{12}{5}x+\frac{36}{25}=\frac{9}{5}+\frac{36}{25}
Square -\frac{6}{5} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{12}{5}x+\frac{36}{25}=\frac{81}{25}
Add \frac{9}{5} to \frac{36}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{6}{5}\right)^{2}=\frac{81}{25}
Factor x^{2}-\frac{12}{5}x+\frac{36}{25}. 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{6}{5}\right)^{2}}=\sqrt{\frac{81}{25}}
Take the square root of both sides of the equation.
x-\frac{6}{5}=\frac{9}{5} x-\frac{6}{5}=-\frac{9}{5}
Simplify.
x=3 x=-\frac{3}{5}
Add \frac{6}{5} to both sides of the equation.
x ^ 2 -\frac{12}{5}x -\frac{9}{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 5
r + s = \frac{12}{5} rs = -\frac{9}{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{6}{5} - u s = \frac{6}{5} + u
Two numbers r and s sum up to \frac{12}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{12}{5} = \frac{6}{5}. 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{6}{5} - u) (\frac{6}{5} + u) = -\frac{9}{5}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{9}{5}
\frac{36}{25} - u^2 = -\frac{9}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{9}{5}-\frac{36}{25} = -\frac{81}{25}
Simplify the expression by subtracting \frac{36}{25} on both sides
u^2 = \frac{81}{25} u = \pm\sqrt{\frac{81}{25}} = \pm \frac{9}{5}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{6}{5} - \frac{9}{5} = -0.600 s = \frac{6}{5} + \frac{9}{5} = 3
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.