3b^2-25b+50=0

Solution for 3b^2-25b+50=0 equation:

3b^2-25b+50=0

a = 3; b = -25; c = +50;
Δ = b2-4ac
Δ = -252-4·3·50
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-5}{2*3}=\frac{20}{6}
=3+1/3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+5}{2*3}=\frac{30}{6}
=5
$