2w^2+17w+21=0

Solution for 2w^2+17w+21=0 equation:

2w^2+17w+21=0

a = 2; b = 17; c = +21;
Δ = b2-4ac
Δ = 172-4·2·21
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{-(17)-11}{2*2}=\frac{-28}{4}
=-7
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+11}{2*2}=\frac{-6}{4}
=-1+1/2
$