2w^2+21w+10=0

Solution for 2w^2+21w+10=0 equation:

2w^2+21w+10=0

a = 2; b = 21; c = +10;
Δ = b2-4ac
Δ = 212-4·2·10
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-19}{2*2}=\frac{-40}{4}
=-10
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+19}{2*2}=\frac{-2}{4}
=-1/2
$