2w^2+13w+6=0

Solution for 2w^2+13w+6=0 equation:

2w^2+13w+6=0

a = 2; b = 13; c = +6;
Δ = b2-4ac
Δ = 132-4·2·6
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-11}{2*2}=\frac{-24}{4}
=-6
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+11}{2*2}=\frac{-2}{4}
=-1/2
$