q^2+6q-55=0

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Solution for q^2+6q-55=0 equation: 



q^2+6q-55=0

a = 1; b = 6; c = -55;
Δ = b2-4ac
Δ = 62-4·1·(-55)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$


$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-16}{2*1}=\frac{-22}{2}
=-11
$


$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+16}{2*1}=\frac{10}{2}
=5
$

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