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3000-99x+0.5x^2=0
a = 0.5; b = -99; c = +3000;
Δ = b2-4ac
Δ = -992-4·0.5·3000
Δ = 3801
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{-(-99)-\sqrt{3801}}{2*0.5}=\frac{99-\sqrt{3801}}{1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-99)+\sqrt{3801}}{2*0.5}=\frac{99+\sqrt{3801}}{1} $
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