6b^2+25b+4=0

Solution for 6b^2+25b+4=0 equation:

6b^2+25b+4=0

a = 6; b = 25; c = +4;
Δ = b2-4ac
Δ = 252-4·6·4
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-23}{2*6}=\frac{-48}{12}
=-4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+23}{2*6}=\frac{-2}{12}
=-1/6
$