26q-0.9q^2-5=0

Solution for 26q-0.9q^2-5=0 equation:

26q-0.9q^2-5=0

a = -0.9; b = 26; c = -5;
Δ = b2-4ac
Δ = 262-4·(-0.9)·(-5)
Δ = 658
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-\sqrt{658}}{2*-0.9}=\frac{-26-\sqrt{658}}{-1.8}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+\sqrt{658}}{2*-0.9}=\frac{-26+\sqrt{658}}{-1.8}
$