6g^2+41g=-70

Solution for 6g^2+41g=-70 equation:

6g^2+41g=-70

We move all terms to the left:

6g^2+41g-(-70)=0

We add all the numbers together, and all the variables

6g^2+41g+70=0

a = 6; b = 41; c = +70;
Δ = b2-4ac
Δ = 412-4·6·70
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-1}{2*6}=\frac{-42}{12}
=-3+1/2
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+1}{2*6}=\frac{-40}{12}
=-3+1/3
$