2w^2+5w-403=0

Solution for 2w^2+5w-403=0 equation:

2w^2+5w-403=0

a = 2; b = 5; c = -403;
Δ = b2-4ac
Δ = 52-4·2·(-403)
Δ = 3249
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3249}=57$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-57}{2*2}=\frac{-62}{4}
=-15+1/2
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+57}{2*2}=\frac{52}{4}
=13
$