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a+b=-14 ab=15\times 3=45
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 15g^{2}+ag+bg+3. To find a and b, set up a system to be solved.
-1,-45 -3,-15 -5,-9
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 45.
-1-45=-46 -3-15=-18 -5-9=-14
Calculate the sum for each pair.
a=-9 b=-5
The solution is the pair that gives sum -14.
\left(15g^{2}-9g\right)+\left(-5g+3\right)
Rewrite 15g^{2}-14g+3 as \left(15g^{2}-9g\right)+\left(-5g+3\right).
3g\left(5g-3\right)-\left(5g-3\right)
Factor out 3g in the first and -1 in the second group.
\left(5g-3\right)\left(3g-1\right)
Factor out common term 5g-3 by using distributive property.
g=\frac{3}{5} g=\frac{1}{3}
To find equation solutions, solve 5g-3=0 and 3g-1=0.
15g^{2}-14g+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.
g=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 15\times 3}}{2\times 15}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 15 for a, -14 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
g=\frac{-\left(-14\right)±\sqrt{196-4\times 15\times 3}}{2\times 15}
Square -14.
g=\frac{-\left(-14\right)±\sqrt{196-60\times 3}}{2\times 15}
Multiply -4 times 15.
g=\frac{-\left(-14\right)±\sqrt{196-180}}{2\times 15}
Multiply -60 times 3.
g=\frac{-\left(-14\right)±\sqrt{16}}{2\times 15}
Add 196 to -180.
g=\frac{-\left(-14\right)±4}{2\times 15}
Take the square root of 16.
g=\frac{14±4}{2\times 15}
The opposite of -14 is 14.
g=\frac{14±4}{30}
Multiply 2 times 15.
g=\frac{18}{30}
Now solve the equation g=\frac{14±4}{30} when ± is plus. Add 14 to 4.
g=\frac{3}{5}
Reduce the fraction \frac{18}{30} to lowest terms by extracting and canceling out 6.
g=\frac{10}{30}
Now solve the equation g=\frac{14±4}{30} when ± is minus. Subtract 4 from 14.
g=\frac{1}{3}
Reduce the fraction \frac{10}{30} to lowest terms by extracting and canceling out 10.
g=\frac{3}{5} g=\frac{1}{3}
The equation is now solved.
15g^{2}-14g+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.
15g^{2}-14g+3-3=-3
Subtract 3 from both sides of the equation.
15g^{2}-14g=-3
Subtracting 3 from itself leaves 0.
\frac{15g^{2}-14g}{15}=-\frac{3}{15}
Divide both sides by 15.
g^{2}-\frac{14}{15}g=-\frac{3}{15}
Dividing by 15 undoes the multiplication by 15.
g^{2}-\frac{14}{15}g=-\frac{1}{5}
Reduce the fraction \frac{-3}{15} to lowest terms by extracting and canceling out 3.
g^{2}-\frac{14}{15}g+\left(-\frac{7}{15}\right)^{2}=-\frac{1}{5}+\left(-\frac{7}{15}\right)^{2}
Divide -\frac{14}{15}, the coefficient of the x term, by 2 to get -\frac{7}{15}. Then add the square of -\frac{7}{15} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
g^{2}-\frac{14}{15}g+\frac{49}{225}=-\frac{1}{5}+\frac{49}{225}
Square -\frac{7}{15} by squaring both the numerator and the denominator of the fraction.
g^{2}-\frac{14}{15}g+\frac{49}{225}=\frac{4}{225}
Add -\frac{1}{5} to \frac{49}{225} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(g-\frac{7}{15}\right)^{2}=\frac{4}{225}
Factor g^{2}-\frac{14}{15}g+\frac{49}{225}. 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(g-\frac{7}{15}\right)^{2}}=\sqrt{\frac{4}{225}}
Take the square root of both sides of the equation.
g-\frac{7}{15}=\frac{2}{15} g-\frac{7}{15}=-\frac{2}{15}
Simplify.
g=\frac{3}{5} g=\frac{1}{3}
Add \frac{7}{15} to both sides of the equation.
x ^ 2 -\frac{14}{15}x +\frac{1}{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 15
r + s = \frac{14}{15} rs = \frac{1}{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{7}{15} - u s = \frac{7}{15} + u
Two numbers r and s sum up to \frac{14}{15} exactly when the average of the two numbers is \frac{1}{2}*\frac{14}{15} = \frac{7}{15}. 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{7}{15} - u) (\frac{7}{15} + u) = \frac{1}{5}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{5}
\frac{49}{225} - u^2 = \frac{1}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{1}{5}-\frac{49}{225} = -\frac{4}{225}
Simplify the expression by subtracting \frac{49}{225} on both sides
u^2 = \frac{4}{225} u = \pm\sqrt{\frac{4}{225}} = \pm \frac{2}{15}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{7}{15} - \frac{2}{15} = 0.333 s = \frac{7}{15} + \frac{2}{15} = 0.600
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.