4y^2+y=39

Solution for 4y^2+y=39 equation:

4y^2+y=39

We move all terms to the left:

4y^2+y-(39)=0

a = 4; b = 1; c = -39;
Δ = b2-4ac
Δ = 12-4·4·(-39)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-25}{2*4}=\frac{-26}{8}
=-3+1/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+25}{2*4}=\frac{24}{8}
=3
$