4q^2+18q-50=0

Solution for 4q^2+18q-50=0 equation:

4q^2+18q-50=0

a = 4; b = 18; c = -50;
Δ = b2-4ac
Δ = 182-4·4·(-50)
Δ = 1124
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:

$\sqrt{\Delta}=\sqrt{1124}=\sqrt{4*281}=\sqrt{4}*\sqrt{281}=2\sqrt{281}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{281}}{2*4}=\frac{-18-2\sqrt{281}}{8}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{281}}{2*4}=\frac{-18+2\sqrt{281}}{8}
$