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0.3x^2+x+0.4=0
a = 0.3; b = 1; c = +0.4;
Δ = b2-4ac
Δ = 12-4·0.3·0.4
Δ = 0.52
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{-(1)-\sqrt{0.52}}{2*0.3}=\frac{-1-\sqrt{0.52}}{0.6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{0.52}}{2*0.3}=\frac{-1+\sqrt{0.52}}{0.6} $
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