-3y^2-28y+196=0

Solution for -3y^2-28y+196=0 equation:

-3y^2-28y+196=0

a = -3; b = -28; c = +196;
Δ = b2-4ac
Δ = -282-4·(-3)·196
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-56}{2*-3}=\frac{-28}{-6}
=4+2/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+56}{2*-3}=\frac{84}{-6}
=-14
$