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8=x^2-4x+3
We move all terms to the left:
8-(x^2-4x+3)=0
We get rid of parentheses
-x^2+4x-3+8=0
We add all the numbers together, and all the variables
-1x^2+4x+5=0
a = -1; b = 4; c = +5;
Δ = b2-4ac
Δ = 42-4·(-1)·5
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6}{2*-1}=\frac{-10}{-2} =+5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6}{2*-1}=\frac{2}{-2} =-1 $
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