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2z^2=z+3
We move all terms to the left:
2z^2-(z+3)=0
We get rid of parentheses
2z^2-z-3=0
We add all the numbers together, and all the variables
2z^2-1z-3=0
a = 2; b = -1; c = -3;
Δ = b2-4ac
Δ = -12-4·2·(-3)
Δ = 25
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{25}=5$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*2}=\frac{-4}{4} =-1 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*2}=\frac{6}{4} =1+1/2 $
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