3z^2+30z+53=-19

Solution for 3z^2+30z+53=-19 equation:

3z^2+30z+53=-19

We move all terms to the left:

3z^2+30z+53-(-19)=0

We add all the numbers together, and all the variables

3z^2+30z+72=0

a = 3; b = 30; c = +72;
Δ = b2-4ac
Δ = 302-4·3·72
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-6}{2*3}=\frac{-36}{6}
=-6
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+6}{2*3}=\frac{-24}{6}
=-4
$