5d^2-22d+9=0

Solution for 5d^2-22d+9=0 equation:

5d^2-22d+9=0

a = 5; b = -22; c = +9;
Δ = b2-4ac
Δ = -222-4·5·9
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-4\sqrt{19}}{2*5}=\frac{22-4\sqrt{19}}{10}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+4\sqrt{19}}{2*5}=\frac{22+4\sqrt{19}}{10}
$