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x^2-4x+2.5=0
a = 1; b = -4; c = +2.5;
Δ = b2-4ac
Δ = -42-4·1·2.5
Δ = 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{6}}{2*1}=\frac{4-\sqrt{6}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+\sqrt{6}}{2*1}=\frac{4+\sqrt{6}}{2} $
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