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0.3x^2+27x-0.2=0
a = 0.3; b = 27; c = -0.2;
Δ = b2-4ac
Δ = 272-4·0.3·(-0.2)
Δ = 729.24
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{-(27)-\sqrt{729.24}}{2*0.3}=\frac{-27-\sqrt{729.24}}{0.6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+\sqrt{729.24}}{2*0.3}=\frac{-27+\sqrt{729.24}}{0.6} $
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