-2w^2+12w+19=0

Solution for -2w^2+12w+19=0 equation:

-2w^2+12w+19=0

a = -2; b = 12; c = +19;
Δ = b2-4ac
Δ = 122-4·(-2)·19
Δ = 296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{296}=\sqrt{4*74}=\sqrt{4}*\sqrt{74}=2\sqrt{74}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{74}}{2*-2}=\frac{-12-2\sqrt{74}}{-4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{74}}{2*-2}=\frac{-12+2\sqrt{74}}{-4}
$