(2w^2+11w+5)=0

Solution for (2w^2+11w+5)=0 equation:

(2w^2+11w+5)=0

We get rid of parentheses

2w^2+11w+5=0

a = 2; b = 11; c = +5;
Δ = b2-4ac
Δ = 112-4·2·5
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-9}{2*2}=\frac{-20}{4}
=-5
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+9}{2*2}=\frac{-2}{4}
=-1/2
$