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2x^2+4x+2x=0
We add all the numbers together, and all the variables
2x^2+6x=0
a = 2; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·2·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*2}=\frac{-12}{4} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*2}=\frac{0}{4} =0 $
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