2q^2-139+20=0

Solution for 2q^2-139+20=0 equation:

2q^2-139+20=0

We add all the numbers together, and all the variables

2q^2-119=0

a = 2; b = 0; c = -119;
Δ = b2-4ac
Δ = 02-4·2·(-119)
Δ = 952
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:

$\sqrt{\Delta}=\sqrt{952}=\sqrt{4*238}=\sqrt{4}*\sqrt{238}=2\sqrt{238}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{238}}{2*2}=\frac{0-2\sqrt{238}}{4}
=-\frac{2\sqrt{238}}{4}
=-\frac{\sqrt{238}}{2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{238}}{2*2}=\frac{0+2\sqrt{238}}{4}
=\frac{2\sqrt{238}}{4}
=\frac{\sqrt{238}}{2}
$