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56z^2+2=27z=0
We move all terms to the left:
56z^2+2-(27z)=0
a = 56; b = -27; c = +2;
Δ = b2-4ac
Δ = -272-4·56·2
Δ = 281
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}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-\sqrt{281}}{2*56}=\frac{27-\sqrt{281}}{112} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+\sqrt{281}}{2*56}=\frac{27+\sqrt{281}}{112} $
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