0=2x2+19x+35

Solution for 0=2x2+19x+35 equation:

0=2x^2+19x+35

We move all terms to the left:

0-(2x^2+19x+35)=0

We add all the numbers together, and all the variables

-(2x^2+19x+35)=0

We get rid of parentheses

-2x^2-19x-35=0

a = -2; b = -19; c = -35;
Δ = b2-4ac
Δ = -192-4·(-2)·(-35)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-9}{2*-2}=\frac{10}{-4}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+9}{2*-2}=\frac{28}{-4}
=-7
$