3w^2+w-52=0

Solution for 3w^2+w-52=0 equation:

3w^2+w-52=0

a = 3; b = 1; c = -52;
Δ = b2-4ac
Δ = 12-4·3·(-52)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-25}{2*3}=\frac{-26}{6}
=-4+1/3
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+25}{2*3}=\frac{24}{6}
=4
$