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3x^2+2x+(=0
We add all the numbers together, and all the variables
3x^2+2x=0
a = 3; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·3·0
Δ = 4
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{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*3}=\frac{-4}{6} =-2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*3}=\frac{0}{6} =0 $
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