J(d)=-5d^2+50d

Solution for J(d)=-5d^2+50d equation:

(J)=-5J^2+50J

We move all terms to the left:

(J)-(-5J^2+50J)=0

We get rid of parentheses

5J^2-50J+J=0

We add all the numbers together, and all the variables

5J^2-49J=0

a = 5; b = -49; c = 0;
Δ = b2-4ac
Δ = -492-4·5·0
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$J_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$J_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$
$J_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-49)-49}{2*5}=\frac{0}{10}
=0
$
$J_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+49}{2*5}=\frac{98}{10}
=9+4/5
$