21z2+32z+12=0

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Solution for 21z2+32z+12=0 equation:



21z^2+32z+12=0
a = 21; b = 32; c = +12;
Δ = b2-4ac
Δ = 322-4·21·12
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4}{2*21}=\frac{-36}{42} =-6/7 $
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4}{2*21}=\frac{-28}{42} =-2/3 $

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