3z2+7z+4=0

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Solution for 3z2+7z+4=0 equation: 



3z^2+7z+4=0

a = 3; b = 7; c = +4;
Δ = b2-4ac
Δ = 72-4·3·4
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{-(7)-1}{2*3}=\frac{-8}{6}
=-1+1/3
$


$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+1}{2*3}=\frac{-6}{6}
=-1
$

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