7(z+2)3z=44

Solution for 7(z+2)3z=44 equation:

7(z+2)3z=44

We move all terms to the left:

7(z+2)3z-(44)=0

We multiply parentheses

21z^2+42z-44=0

a = 21; b = 42; c = -44;
Δ = b2-4ac
Δ = 422-4·21·(-44)
Δ = 5460
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{5460}=\sqrt{4*1365}=\sqrt{4}*\sqrt{1365}=2\sqrt{1365}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{1365}}{2*21}=\frac{-42-2\sqrt{1365}}{42}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{1365}}{2*21}=\frac{-42+2\sqrt{1365}}{42}
$