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0.1x^2+2x-150=0
a = 0.1; b = 2; c = -150;
Δ = b2-4ac
Δ = 22-4·0.1·(-150)
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*0.1}=\frac{-10}{0.2} =-50 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*0.1}=\frac{6}{0.2} =30 $
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