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0=(-2x+3)(-x+3)
We move all terms to the left:
0-((-2x+3)(-x+3))=0
We add all the numbers together, and all the variables
-((-2x+3)(-1x+3))+0=0
We add all the numbers together, and all the variables
-((-2x+3)(-1x+3))=0
We multiply parentheses ..
-((+2x^2-6x-3x+9))=0
We calculate terms in parentheses: -((+2x^2-6x-3x+9)), so:We get rid of parentheses
(+2x^2-6x-3x+9)
We get rid of parentheses
2x^2-6x-3x+9
We add all the numbers together, and all the variables
2x^2-9x+9
Back to the equation:
-(2x^2-9x+9)
-2x^2+9x-9=0
a = -2; b = 9; c = -9;
Δ = b2-4ac
Δ = 92-4·(-2)·(-9)
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3}{2*-2}=\frac{-12}{-4} =+3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3}{2*-2}=\frac{-6}{-4} =1+1/2 $
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