10w^2+17w+3=0

Solution for 10w^2+17w+3=0 equation:

10w^2+17w+3=0

a = 10; b = 17; c = +3;
Δ = b2-4ac
Δ = 172-4·10·3
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-13}{2*10}=\frac{-30}{20}
=-1+1/2
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+13}{2*10}=\frac{-4}{20}
=-1/5
$