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x^2-5x+4=03
We move all terms to the left:
x^2-5x+4-(03)=0
We add all the numbers together, and all the variables
x^2-5x+1=0
a = 1; b = -5; c = +1;
Δ = b2-4ac
Δ = -52-4·1·1
Δ = 21
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{-(-5)-\sqrt{21}}{2*1}=\frac{5-\sqrt{21}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{21}}{2*1}=\frac{5+\sqrt{21}}{2} $
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