2x^2+3x-19=108

Solution for 2x^2+3x-19=108 equation:

2x^2+3x-19=108

We move all terms to the left:

2x^2+3x-19-(108)=0

We add all the numbers together, and all the variables

2x^2+3x-127=0

a = 2; b = 3; c = -127;
Δ = b2-4ac
Δ = 32-4·2·(-127)
Δ = 1025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1025}=\sqrt{25*41}=\sqrt{25}*\sqrt{41}=5\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-5\sqrt{41}}{2*2}=\frac{-3-5\sqrt{41}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+5\sqrt{41}}{2*2}=\frac{-3+5\sqrt{41}}{4}
$