33000/3z+50+z-20=300

Solution for 33000/3z+50+z-20=300 equation:

33000/3z+50+z-20=300

We move all terms to the left:

33000/3z+50+z-20-(300)=0


Domain of the equation: 3z!=0

z!=0/3

z!=0

z∈R


We add all the numbers together, and all the variables

z+33000/3z-270=0

We multiply all the terms by the denominator


z*3z-270*3z+33000=0

Wy multiply elements

3z^2-810z+33000=0

a = 3; b = -810; c = +33000;
Δ = b2-4ac
Δ = -8102-4·3·33000
Δ = 260100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{260100}=510$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-810)-510}{2*3}=\frac{300}{6}
=50
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-810)+510}{2*3}=\frac{1320}{6}
=220
$