10b^2+29b=21

Solution for 10b^2+29b=21 equation:

10b^2+29b=21

We move all terms to the left:

10b^2+29b-(21)=0

a = 10; b = 29; c = -21;
Δ = b2-4ac
Δ = 292-4·10·(-21)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-41}{2*10}=\frac{-70}{20}
=-3+1/2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+41}{2*10}=\frac{12}{20}
=3/5
$