14+g(4g-3)=40

Solution for 14+g(4g-3)=40 equation:

14+g(4g-3)=40

We move all terms to the left:

14+g(4g-3)-(40)=0

We add all the numbers together, and all the variables

g(4g-3)-26=0

We multiply parentheses

4g^2-3g-26=0

a = 4; b = -3; c = -26;
Δ = b2-4ac
Δ = -32-4·4·(-26)
Δ = 425
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:

$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5\sqrt{17}}{2*4}=\frac{3-5\sqrt{17}}{8}
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5\sqrt{17}}{2*4}=\frac{3+5\sqrt{17}}{8}
$