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x^2-4x+3.05=0
a = 1; b = -4; c = +3.05;
Δ = b2-4ac
Δ = -42-4·1·3.05
Δ = 3.8
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{3.8}}{2*1}=\frac{4-\sqrt{3.8}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+\sqrt{3.8}}{2*1}=\frac{4+\sqrt{3.8}}{2} $
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