-1.8q^2+4q+20=0

Solution for -1.8q^2+4q+20=0 equation:

-1.8q^2+4q+20=0

a = -1.8; b = 4; c = +20;
Δ = b2-4ac
Δ = 42-4·(-1.8)·20
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{10}}{2*-1.8}=\frac{-4-4\sqrt{10}}{-3.6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{10}}{2*-1.8}=\frac{-4+4\sqrt{10}}{-3.6}
$