3n^2+7n=98

Solution for 3n^2+7n=98 equation:

3n^2+7n=98

We move all terms to the left:

3n^2+7n-(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:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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