3d^2+8d=40

Solution for 3d^2+8d=40 equation:

3d^2+8d=40

We move all terms to the left:

3d^2+8d-(40)=0

a = 3; b = 8; c = -40;
Δ = b2-4ac
Δ = 82-4·3·(-40)
Δ = 544
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:

$\sqrt{\Delta}=\sqrt{544}=\sqrt{16*34}=\sqrt{16}*\sqrt{34}=4\sqrt{34}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{34}}{2*3}=\frac{-8-4\sqrt{34}}{6}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{34}}{2*3}=\frac{-8+4\sqrt{34}}{6}
$