6b^2+29b+35=0

Solution for 6b^2+29b+35=0 equation:

6b^2+29b+35=0

a = 6; b = 29; c = +35;
Δ = b2-4ac
Δ = 292-4·6·35
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-1}{2*6}=\frac{-30}{12}
=-2+1/2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+1}{2*6}=\frac{-28}{12}
=-2+1/3
$