16q^2+200q-45=0

Solution for 16q^2+200q-45=0 equation:

16q^2+200q-45=0

a = 16; b = 200; c = -45;
Δ = b2-4ac
Δ = 2002-4·16·(-45)
Δ = 42880
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:

$\sqrt{\Delta}=\sqrt{42880}=\sqrt{64*670}=\sqrt{64}*\sqrt{670}=8\sqrt{670}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-8\sqrt{670}}{2*16}=\frac{-200-8\sqrt{670}}{32}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+8\sqrt{670}}{2*16}=\frac{-200+8\sqrt{670}}{32}
$