21n^2+32n+12=0

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Solution for 21n^2+32n+12=0 equation: 



21n^2+32n+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:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


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


$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4}{2*21}=\frac{-28}{42}
=-2/3
$

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