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-2850+149x+0.1x^2=0
a = 0.1; b = 149; c = -2850;
Δ = b2-4ac
Δ = 1492-4·0.1·(-2850)
Δ = 23341
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{-(149)-\sqrt{23341}}{2*0.1}=\frac{-149-\sqrt{23341}}{0.2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(149)+\sqrt{23341}}{2*0.1}=\frac{-149+\sqrt{23341}}{0.2} $
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