m^2+18m+77=0

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Solution for m^2+18m+77=0 equation: 



m^2+18m+77=0

a = 1; b = 18; c = +77;
Δ = b2-4ac
Δ = 182-4·1·77
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-4}{2*1}=\frac{-22}{2}
=-11
$


$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+4}{2*1}=\frac{-14}{2}
=-7
$

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