3b^2+7b-98=0

Solution for 3b^2+7b-98=0 equation:

3b^2+7b-98=0

a = 3; b = 7; c = -98;
Δ = b2-4ac
Δ = 72-4·3·(-98)
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-35}{2*3}=\frac{-42}{6}
=-7
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+35}{2*3}=\frac{28}{6}
=4+2/3
$