-3e^2+7e+6=0

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Solution for -3e^2+7e+6=0 equation:



-3e^2+7e+6=0
a = -3; b = 7; c = +6;
Δ = b2-4ac
Δ = 72-4·(-3)·6
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-11}{2*-3}=\frac{-18}{-6} =+3 $
$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+11}{2*-3}=\frac{4}{-6} =-2/3 $

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