3w^2+5w-492=0

Solution for 3w^2+5w-492=0 equation:

3w^2+5w-492=0

a = 3; b = 5; c = -492;
Δ = b2-4ac
Δ = 52-4·3·(-492)
Δ = 5929
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5929}=77$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-77}{2*3}=\frac{-82}{6}
=-13+2/3
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+77}{2*3}=\frac{72}{6}
=12
$