4w^2+25w+6=0

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

4w^2+25w+6=0

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

$\sqrt{\Delta}=\sqrt{529}=23$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-23}{2*4}=\frac{-48}{8}
=-6
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+23}{2*4}=\frac{-2}{8}
=-1/4
$