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0.05x^2-4x-38=0
a = 0.05; b = -4; c = -38;
Δ = b2-4ac
Δ = -42-4·0.05·(-38)
Δ = 23.6
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{-(-4)-\sqrt{23.6}}{2*0.05}=\frac{4-\sqrt{23.6}}{0.1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+\sqrt{23.6}}{2*0.05}=\frac{4+\sqrt{23.6}}{0.1} $
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