7a^2+13a+6=0

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Solution for 7a^2+13a+6=0 equation: 



7a^2+13a+6=0

a = 7; b = 13; c = +6;
Δ = b2-4ac
Δ = 132-4·7·6
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$


$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-1}{2*7}=\frac{-14}{14}
=-1
$


$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+1}{2*7}=\frac{-12}{14}
=-6/7
$

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