2u2+8u+6=0

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Solution for 2u2+8u+6=0 equation: 



2u^2+8u+6=0

a = 2; b = 8; c = +6;
Δ = b2-4ac
Δ = 82-4·2·6
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4}{2*2}=\frac{-12}{4}
=-3
$


$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*2}=\frac{-4}{4}
=-1
$

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