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2a^2+6a+4=0
a = 2; b = 6; c = +4;
Δ = b2-4ac
Δ = 62-4·2·4
Δ = 4
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{4}=2$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2}{2*2}=\frac{-8}{4} =-2 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2}{2*2}=\frac{-4}{4} =-1 $
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