Solve for d
d=\frac{1}{3}\approx 0.333333333
d = \frac{7}{5} = 1\frac{2}{5} = 1.4
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15d^{2}-26d+7=0
Divide both sides by 2.
a+b=-26 ab=15\times 7=105
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 15d^{2}+ad+bd+7. To find a and b, set up a system to be solved.
-1,-105 -3,-35 -5,-21 -7,-15
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 105.
-1-105=-106 -3-35=-38 -5-21=-26 -7-15=-22
Calculate the sum for each pair.
a=-21 b=-5
The solution is the pair that gives sum -26.
\left(15d^{2}-21d\right)+\left(-5d+7\right)
Rewrite 15d^{2}-26d+7 as \left(15d^{2}-21d\right)+\left(-5d+7\right).
3d\left(5d-7\right)-\left(5d-7\right)
Factor out 3d in the first and -1 in the second group.
\left(5d-7\right)\left(3d-1\right)
Factor out common term 5d-7 by using distributive property.
d=\frac{7}{5} d=\frac{1}{3}
To find equation solutions, solve 5d-7=0 and 3d-1=0.
30d^{2}-52d+14=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.
d=\frac{-\left(-52\right)±\sqrt{\left(-52\right)^{2}-4\times 30\times 14}}{2\times 30}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 30 for a, -52 for b, and 14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
d=\frac{-\left(-52\right)±\sqrt{2704-4\times 30\times 14}}{2\times 30}
Square -52.
d=\frac{-\left(-52\right)±\sqrt{2704-120\times 14}}{2\times 30}
Multiply -4 times 30.
d=\frac{-\left(-52\right)±\sqrt{2704-1680}}{2\times 30}
Multiply -120 times 14.
d=\frac{-\left(-52\right)±\sqrt{1024}}{2\times 30}
Add 2704 to -1680.
d=\frac{-\left(-52\right)±32}{2\times 30}
Take the square root of 1024.
d=\frac{52±32}{2\times 30}
The opposite of -52 is 52.
d=\frac{52±32}{60}
Multiply 2 times 30.
d=\frac{84}{60}
Now solve the equation d=\frac{52±32}{60} when ± is plus. Add 52 to 32.
d=\frac{7}{5}
Reduce the fraction \frac{84}{60} to lowest terms by extracting and canceling out 12.
d=\frac{20}{60}
Now solve the equation d=\frac{52±32}{60} when ± is minus. Subtract 32 from 52.
d=\frac{1}{3}
Reduce the fraction \frac{20}{60} to lowest terms by extracting and canceling out 20.
d=\frac{7}{5} d=\frac{1}{3}
The equation is now solved.
30d^{2}-52d+14=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.
30d^{2}-52d+14-14=-14
Subtract 14 from both sides of the equation.
30d^{2}-52d=-14
Subtracting 14 from itself leaves 0.
\frac{30d^{2}-52d}{30}=-\frac{14}{30}
Divide both sides by 30.
d^{2}+\left(-\frac{52}{30}\right)d=-\frac{14}{30}
Dividing by 30 undoes the multiplication by 30.
d^{2}-\frac{26}{15}d=-\frac{14}{30}
Reduce the fraction \frac{-52}{30} to lowest terms by extracting and canceling out 2.
d^{2}-\frac{26}{15}d=-\frac{7}{15}
Reduce the fraction \frac{-14}{30} to lowest terms by extracting and canceling out 2.
d^{2}-\frac{26}{15}d+\left(-\frac{13}{15}\right)^{2}=-\frac{7}{15}+\left(-\frac{13}{15}\right)^{2}
Divide -\frac{26}{15}, the coefficient of the x term, by 2 to get -\frac{13}{15}. Then add the square of -\frac{13}{15} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
d^{2}-\frac{26}{15}d+\frac{169}{225}=-\frac{7}{15}+\frac{169}{225}
Square -\frac{13}{15} by squaring both the numerator and the denominator of the fraction.
d^{2}-\frac{26}{15}d+\frac{169}{225}=\frac{64}{225}
Add -\frac{7}{15} to \frac{169}{225} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(d-\frac{13}{15}\right)^{2}=\frac{64}{225}
Factor d^{2}-\frac{26}{15}d+\frac{169}{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(d-\frac{13}{15}\right)^{2}}=\sqrt{\frac{64}{225}}
Take the square root of both sides of the equation.
d-\frac{13}{15}=\frac{8}{15} d-\frac{13}{15}=-\frac{8}{15}
Simplify.
d=\frac{7}{5} d=\frac{1}{3}
Add \frac{13}{15} to both sides of the equation.
x ^ 2 -\frac{26}{15}x +\frac{7}{15} = 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 30
r + s = \frac{26}{15} rs = \frac{7}{15}
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{13}{15} - u s = \frac{13}{15} + u
Two numbers r and s sum up to \frac{26}{15} exactly when the average of the two numbers is \frac{1}{2}*\frac{26}{15} = \frac{13}{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{13}{15} - u) (\frac{13}{15} + u) = \frac{7}{15}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{7}{15}
\frac{169}{225} - u^2 = \frac{7}{15}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{7}{15}-\frac{169}{225} = -\frac{64}{225}
Simplify the expression by subtracting \frac{169}{225} on both sides
u^2 = \frac{64}{225} u = \pm\sqrt{\frac{64}{225}} = \pm \frac{8}{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{13}{15} - \frac{8}{15} = 0.333 s = \frac{13}{15} + \frac{8}{15} = 1.400
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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