2z^2+5z-7=0

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Solution for 2z^2+5z-7=0 equation: 



2z^2+5z-7=0

a = 2; b = 5; c = -7;
Δ = b2-4ac
Δ = 52-4·2·(-7)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$


$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-9}{2*2}=\frac{-14}{4}
=-3+1/2
$


$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+9}{2*2}=\frac{4}{4}
=1
$

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