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0=(x-2)(x-3)
We move all terms to the left:
0-((x-2)(x-3))=0
We add all the numbers together, and all the variables
-((x-2)(x-3))=0
We multiply parentheses ..
-((+x^2-3x-2x+6))=0
We calculate terms in parentheses: -((+x^2-3x-2x+6)), so:We get rid of parentheses
(+x^2-3x-2x+6)
We get rid of parentheses
x^2-3x-2x+6
We add all the numbers together, and all the variables
x^2-5x+6
Back to the equation:
-(x^2-5x+6)
-x^2+5x-6=0
We add all the numbers together, and all the variables
-1x^2+5x-6=0
a = -1; b = 5; c = -6;
Δ = b2-4ac
Δ = 52-4·(-1)·(-6)
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-1}{2*-1}=\frac{-6}{-2} =+3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+1}{2*-1}=\frac{-4}{-2} =+2 $
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