0.5q^2+4q+1=0

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Solution for 0.5q^2+4q+1=0 equation: 



0.5q^2+4q+1=0

a = 0.5; b = 4; c = +1;
Δ = b2-4ac
Δ = 42-4·0.5·1
Δ = 14
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-\sqrt{14}}{2*0.5}=\frac{-4-\sqrt{14}}{1}
$


$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+\sqrt{14}}{2*0.5}=\frac{-4+\sqrt{14}}{1}
$

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