c^2+16c+63=0

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Solution for c^2+16c+63=0 equation: 



c^2+16c+63=0

a = 1; b = 16; c = +63;
Δ = b2-4ac
Δ = 162-4·1·63
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{-(16)-2}{2*1}=\frac{-18}{2}
=-9
$


$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2}{2*1}=\frac{-14}{2}
=-7
$

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