3b^2+42=25b

Solution for 3b^2+42=25b equation:

3b^2+42=25b

We move all terms to the left:

3b^2+42-(25b)=0

a = 3; b = -25; c = +42;
Δ = b2-4ac
Δ = -252-4·3·42
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-11}{2*3}=\frac{14}{6}
=2+1/3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+11}{2*3}=\frac{36}{6}
=6
$