2d^2+11d+20=d^2

Solution for 2d^2+11d+20=d^2 equation:

2d^2+11d+20=d^2

We move all terms to the left:

2d^2+11d+20-(d^2)=0

determiningTheFunctionDomain
2d^2-d^2+11d+20=0

We add all the numbers together, and all the variables

d^2+11d+20=0

a = 1; b = 11; c = +20;
Δ = b2-4ac
Δ = 112-4·1·20
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{41}}{2*1}=\frac{-11-\sqrt{41}}{2}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{41}}{2*1}=\frac{-11+\sqrt{41}}{2}
$