Skip to main content
Solve for d
Tick mark Image

Similar Problems from Web Search

Share

a+b=-11 ab=6\left(-10\right)=-60
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6d^{2}+ad+bd-10. To find a and b, set up a system to be solved.
1,-60 2,-30 3,-20 4,-15 5,-12 6,-10
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 -60.
1-60=-59 2-30=-28 3-20=-17 4-15=-11 5-12=-7 6-10=-4
Calculate the sum for each pair.
a=-15 b=4
The solution is the pair that gives sum -11.
\left(6d^{2}-15d\right)+\left(4d-10\right)
Rewrite 6d^{2}-11d-10 as \left(6d^{2}-15d\right)+\left(4d-10\right).
3d\left(2d-5\right)+2\left(2d-5\right)
Factor out 3d in the first and 2 in the second group.
\left(2d-5\right)\left(3d+2\right)
Factor out common term 2d-5 by using distributive property.
d=\frac{5}{2} d=-\frac{2}{3}
To find equation solutions, solve 2d-5=0 and 3d+2=0.
6d^{2}-11d-10=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(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 6\left(-10\right)}}{2\times 6}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -11 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
d=\frac{-\left(-11\right)±\sqrt{121-4\times 6\left(-10\right)}}{2\times 6}
Square -11.
d=\frac{-\left(-11\right)±\sqrt{121-24\left(-10\right)}}{2\times 6}
Multiply -4 times 6.
d=\frac{-\left(-11\right)±\sqrt{121+240}}{2\times 6}
Multiply -24 times -10.
d=\frac{-\left(-11\right)±\sqrt{361}}{2\times 6}
Add 121 to 240.
d=\frac{-\left(-11\right)±19}{2\times 6}
Take the square root of 361.
d=\frac{11±19}{2\times 6}
The opposite of -11 is 11.
d=\frac{11±19}{12}
Multiply 2 times 6.
d=\frac{30}{12}
Now solve the equation d=\frac{11±19}{12} when ± is plus. Add 11 to 19.
d=\frac{5}{2}
Reduce the fraction \frac{30}{12} to lowest terms by extracting and canceling out 6.
d=-\frac{8}{12}
Now solve the equation d=\frac{11±19}{12} when ± is minus. Subtract 19 from 11.
d=-\frac{2}{3}
Reduce the fraction \frac{-8}{12} to lowest terms by extracting and canceling out 4.
d=\frac{5}{2} d=-\frac{2}{3}
The equation is now solved.
6d^{2}-11d-10=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.
6d^{2}-11d-10-\left(-10\right)=-\left(-10\right)
Add 10 to both sides of the equation.
6d^{2}-11d=-\left(-10\right)
Subtracting -10 from itself leaves 0.
6d^{2}-11d=10
Subtract -10 from 0.
\frac{6d^{2}-11d}{6}=\frac{10}{6}
Divide both sides by 6.
d^{2}-\frac{11}{6}d=\frac{10}{6}
Dividing by 6 undoes the multiplication by 6.
d^{2}-\frac{11}{6}d=\frac{5}{3}
Reduce the fraction \frac{10}{6} to lowest terms by extracting and canceling out 2.
d^{2}-\frac{11}{6}d+\left(-\frac{11}{12}\right)^{2}=\frac{5}{3}+\left(-\frac{11}{12}\right)^{2}
Divide -\frac{11}{6}, the coefficient of the x term, by 2 to get -\frac{11}{12}. Then add the square of -\frac{11}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
d^{2}-\frac{11}{6}d+\frac{121}{144}=\frac{5}{3}+\frac{121}{144}
Square -\frac{11}{12} by squaring both the numerator and the denominator of the fraction.
d^{2}-\frac{11}{6}d+\frac{121}{144}=\frac{361}{144}
Add \frac{5}{3} to \frac{121}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(d-\frac{11}{12}\right)^{2}=\frac{361}{144}
Factor d^{2}-\frac{11}{6}d+\frac{121}{144}. 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{11}{12}\right)^{2}}=\sqrt{\frac{361}{144}}
Take the square root of both sides of the equation.
d-\frac{11}{12}=\frac{19}{12} d-\frac{11}{12}=-\frac{19}{12}
Simplify.
d=\frac{5}{2} d=-\frac{2}{3}
Add \frac{11}{12} to both sides of the equation.
x ^ 2 -\frac{11}{6}x -\frac{5}{3} = 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 6
r + s = \frac{11}{6} rs = -\frac{5}{3}
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{11}{12} - u s = \frac{11}{12} + u
Two numbers r and s sum up to \frac{11}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{11}{6} = \frac{11}{12}. 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{11}{12} - u) (\frac{11}{12} + u) = -\frac{5}{3}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{3}
\frac{121}{144} - u^2 = -\frac{5}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{5}{3}-\frac{121}{144} = -\frac{361}{144}
Simplify the expression by subtracting \frac{121}{144} on both sides
u^2 = \frac{361}{144} u = \pm\sqrt{\frac{361}{144}} = \pm \frac{19}{12}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{11}{12} - \frac{19}{12} = -0.667 s = \frac{11}{12} + \frac{19}{12} = 2.500
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.