5d^2+21d+9=d+1

Solution for 5d^2+21d+9=d+1 equation:

5d^2+21d+9=d+1

We move all terms to the left:

5d^2+21d+9-(d+1)=0

We get rid of parentheses

5d^2+21d-d-1+9=0

We add all the numbers together, and all the variables

5d^2+20d+8=0

a = 5; b = 20; c = +8;
Δ = b2-4ac
Δ = 202-4·5·8
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{15}}{2*5}=\frac{-20-4\sqrt{15}}{10}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{15}}{2*5}=\frac{-20+4\sqrt{15}}{10}
$