0=2x^2+10x-19

Solution for 0=2x^2+10x-19 equation:

0=2x^2+10x-19

We move all terms to the left:

0-(2x^2+10x-19)=0

We add all the numbers together, and all the variables

-(2x^2+10x-19)=0

We get rid of parentheses

-2x^2-10x+19=0

a = -2; b = -10; c = +19;
Δ = b2-4ac
Δ = -102-4·(-2)·19
Δ = 252
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:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-6\sqrt{7}}{2*-2}=\frac{10-6\sqrt{7}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+6\sqrt{7}}{2*-2}=\frac{10+6\sqrt{7}}{-4}
$