-3d^2+8d^2+16=20d+20

Solution for -3d^2+8d^2+16=20d+20 equation:

-3d^2+8d^2+16=20d+20

We move all terms to the left:

-3d^2+8d^2+16-(20d+20)=0

We add all the numbers together, and all the variables

5d^2-(20d+20)+16=0

We get rid of parentheses

5d^2-20d-20+16=0

We add all the numbers together, and all the variables

5d^2-20d-4=0

a = 5; b = -20; c = -4;
Δ = b2-4ac
Δ = -202-4·5·(-4)
Δ = 480
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:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{30}}{2*5}=\frac{20-4\sqrt{30}}{10}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{30}}{2*5}=\frac{20+4\sqrt{30}}{10}
$