2q^2+7q+1=0

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Solution for 2q^2+7q+1=0 equation: 



2q^2+7q+1=0

a = 2; b = 7; c = +1;
Δ = b2-4ac
Δ = 72-4·2·1
Δ = 41
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{-(7)-\sqrt{41}}{2*2}=\frac{-7-\sqrt{41}}{4}
$


$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{41}}{2*2}=\frac{-7+\sqrt{41}}{4}
$

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