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(5/2)x+(x-4)/2=11
We move all terms to the left:
(5/2)x+(x-4)/2-(11)=0
Domain of the equation: 2)x!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
(+5/2)x+(x-4)/2-11=0
We multiply parentheses
5x^2+(x-4)/2-11=0
We multiply all the terms by the denominator
5x^2*2+(x-4)-11*2=0
We add all the numbers together, and all the variables
5x^2*2+(x-4)-22=0
Wy multiply elements
10x^2+(x-4)-22=0
We get rid of parentheses
10x^2+x-4-22=0
We add all the numbers together, and all the variables
10x^2+x-26=0
a = 10; b = 1; c = -26;
Δ = b2-4ac
Δ = 12-4·10·(-26)
Δ = 1041
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{1041}}{2*10}=\frac{-1-\sqrt{1041}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1041}}{2*10}=\frac{-1+\sqrt{1041}}{20} $
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