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3x-5=x^2-6x+9
We move all terms to the left:
3x-5-(x^2-6x+9)=0
We get rid of parentheses
-x^2+3x+6x-9-5=0
We add all the numbers together, and all the variables
-1x^2+9x-14=0
a = -1; b = 9; c = -14;
Δ = b2-4ac
Δ = 92-4·(-1)·(-14)
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-5}{2*-1}=\frac{-14}{-2} =+7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+5}{2*-1}=\frac{-4}{-2} =+2 $
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