1g^2+18g=0

Solution for 1g^2+18g=0 equation:

1g^2+18g=0

We add all the numbers together, and all the variables

g^2+18g=0

a = 1; b = 18; c = 0;
Δ = b2-4ac
Δ = 182-4·1·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-18}{2*1}=\frac{-36}{2}
=-18
$
$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+18}{2*1}=\frac{0}{2}
=0
$