d^2-36d+320=0

Solution for d^2-36d+320=0 equation:

d^2-36d+320=0

a = 1; b = -36; c = +320;
Δ = b2-4ac
Δ = -362-4·1·320
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-4}{2*1}=\frac{32}{2}
=16
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4}{2*1}=\frac{40}{2}
=20
$