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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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