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0=(2x-3)(3x-5)
We move all terms to the left:
0-((2x-3)(3x-5))=0
We add all the numbers together, and all the variables
-((2x-3)(3x-5))=0
We multiply parentheses ..
-((+6x^2-10x-9x+15))=0
We calculate terms in parentheses: -((+6x^2-10x-9x+15)), so:We get rid of parentheses
(+6x^2-10x-9x+15)
We get rid of parentheses
6x^2-10x-9x+15
We add all the numbers together, and all the variables
6x^2-19x+15
Back to the equation:
-(6x^2-19x+15)
-6x^2+19x-15=0
a = -6; b = 19; c = -15;
Δ = b2-4ac
Δ = 192-4·(-6)·(-15)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-1}{2*-6}=\frac{-20}{-12} =1+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+1}{2*-6}=\frac{-18}{-12} =1+1/2 $
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