12c+c^2+35=0

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Solution for 12c+c^2+35=0 equation: 



12c+c^2+35=0

a = 1; b = 12; c = +35;
Δ = b2-4ac
Δ = 122-4·1·35
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$


$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2}{2*1}=\frac{-14}{2}
=-7
$


$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2}{2*1}=\frac{-10}{2}
=-5
$

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