14z^2-14z=28

Solution for 14z^2-14z=28 equation:

14z^2-14z=28

We move all terms to the left:

14z^2-14z-(28)=0

a = 14; b = -14; c = -28;
Δ = b2-4ac
Δ = -142-4·14·(-28)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-42}{2*14}=\frac{-28}{28}
=-1
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+42}{2*14}=\frac{56}{28}
=2
$