10b^2+29b+4=-6

Solution for 10b^2+29b+4=-6 equation:

10b^2+29b+4=-6

We move all terms to the left:

10b^2+29b+4-(-6)=0

We add all the numbers together, and all the variables

10b^2+29b+10=0

a = 10; b = 29; c = +10;
Δ = b2-4ac
Δ = 292-4·10·10
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-21}{2*10}=\frac{-50}{20}
=-2+1/2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+21}{2*10}=\frac{-8}{20}
=-2/5
$