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1.2x^2+x=0
a = 1.2; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·1.2·0
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*1.2}=\frac{-2}{2.4} =-1/1.2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*1.2}=\frac{0}{2.4} =0 $
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