4z^2+9z+5=0

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Solution for 4z^2+9z+5=0 equation: 



4z^2+9z+5=0

a = 4; b = 9; c = +5;
Δ = b2-4ac
Δ = 92-4·4·5
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$


$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-1}{2*4}=\frac{-10}{8}
=-1+1/4
$


$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+1}{2*4}=\frac{-8}{8}
=-1
$

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