-4(3g-5)10g=19

Solution for -4(3g-5)10g=19 equation:

-4(3g-5)10g=19

We move all terms to the left:

-4(3g-5)10g-(19)=0

We multiply parentheses

-120g^2+200g-19=0

a = -120; b = 200; c = -19;
Δ = b2-4ac
Δ = 2002-4·(-120)·(-19)
Δ = 30880
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{30880}=\sqrt{16*1930}=\sqrt{16}*\sqrt{1930}=4\sqrt{1930}$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-4\sqrt{1930}}{2*-120}=\frac{-200-4\sqrt{1930}}{-240}
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+4\sqrt{1930}}{2*-120}=\frac{-200+4\sqrt{1930}}{-240}
$