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2x^2-7x+1=0
a = 2; b = -7; c = +1;
Δ = b2-4ac
Δ = -72-4·2·1
Δ = 41
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{-(-7)-\sqrt{41}}{2*2}=\frac{7-\sqrt{41}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{41}}{2*2}=\frac{7+\sqrt{41}}{4} $
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