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x^2-4x+20=17
We move all terms to the left:
x^2-4x+20-(17)=0
We add all the numbers together, and all the variables
x^2-4x+3=0
a = 1; b = -4; c = +3;
Δ = b2-4ac
Δ = -42-4·1·3
Δ = 4
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2}{2*1}=\frac{2}{2} =1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2}{2*1}=\frac{6}{2} =3 $
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