3d^2-40d+56=0

Solution for 3d^2-40d+56=0 equation:

3d^2-40d+56=0

a = 3; b = -40; c = +56;
Δ = b2-4ac
Δ = -402-4·3·56
Δ = 928
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{928}=\sqrt{16*58}=\sqrt{16}*\sqrt{58}=4\sqrt{58}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{58}}{2*3}=\frac{40-4\sqrt{58}}{6}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{58}}{2*3}=\frac{40+4\sqrt{58}}{6}
$