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X^2-4X+3=2X-2
We move all terms to the left:
X^2-4X+3-(2X-2)=0
We get rid of parentheses
X^2-4X-2X+2+3=0
We add all the numbers together, and all the variables
X^2-6X+5=0
a = 1; b = -6; c = +5;
Δ = b2-4ac
Δ = -62-4·1·5
Δ = 16
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{16}=4$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-4}{2*1}=\frac{2}{2} =1 $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+4}{2*1}=\frac{10}{2} =5 $
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