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0.05x^2+49.95x-50=0
a = 0.05; b = 49.95; c = -50;
Δ = b2-4ac
Δ = 49.952-4·0.05·(-50)
Δ = 2505.0025
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{-(49.95)-\sqrt{2505.0025}}{2*0.05}=\frac{-49.95-\sqrt{2505.0025}}{0.1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49.95)+\sqrt{2505.0025}}{2*0.05}=\frac{-49.95+\sqrt{2505.0025}}{0.1} $
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