6b^2-29b+20=0

Solution for 6b^2-29b+20=0 equation:

6b^2-29b+20=0

a = 6; b = -29; c = +20;
Δ = b2-4ac
Δ = -292-4·6·20
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-19}{2*6}=\frac{10}{12}
=5/6
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+19}{2*6}=\frac{48}{12}
=4
$