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0.3x^2+0.5x-1.9=0
a = 0.3; b = 0.5; c = -1.9;
Δ = b2-4ac
Δ = 0.52-4·0.3·(-1.9)
Δ = 2.53
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.5)-\sqrt{2.53}}{2*0.3}=\frac{-0.5-\sqrt{2.53}}{0.6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{2.53}}{2*0.3}=\frac{-0.5+\sqrt{2.53}}{0.6} $
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