3z^2+20z+1=8

Solution for 3z^2+20z+1=8 equation:

3z^2+20z+1=8

We move all terms to the left:

3z^2+20z+1-(8)=0

We add all the numbers together, and all the variables

3z^2+20z-7=0

a = 3; b = 20; c = -7;
Δ = b2-4ac
Δ = 202-4·3·(-7)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-22}{2*3}=\frac{-42}{6}
=-7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+22}{2*3}=\frac{2}{6}
=1/3
$