5q^2+q-1=0

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Solution for 5q^2+q-1=0 equation: 



5q^2+q-1=0

a = 5; b = 1; c = -1;
Δ = b2-4ac
Δ = 12-4·5·(-1)
Δ = 21
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{-(1)-\sqrt{21}}{2*5}=\frac{-1-\sqrt{21}}{10}
$


$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{21}}{2*5}=\frac{-1+\sqrt{21}}{10}
$

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