g^2+7g=98

Solution for g^2+7g=98 equation:

g^2+7g=98

We move all terms to the left:

g^2+7g-(98)=0

a = 1; b = 7; c = -98;
Δ = b2-4ac
Δ = 72-4·1·(-98)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-21}{2*1}=\frac{-28}{2}
=-14
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+21}{2*1}=\frac{14}{2}
=7
$