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(z-3)(2z+1)=0
We multiply parentheses ..
(+2z^2+z-6z-3)=0
We get rid of parentheses
2z^2+z-6z-3=0
We add all the numbers together, and all the variables
2z^2-5z-3=0
a = 2; b = -5; c = -3;
Δ = b2-4ac
Δ = -52-4·2·(-3)
Δ = 49
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{49}=7$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-7}{2*2}=\frac{-2}{4} =-1/2 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+7}{2*2}=\frac{12}{4} =3 $
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