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(1x/2)+7-(1x/4)=(19/2)
We move all terms to the left:
(1x/2)+7-(1x/4)-((19/2))=0
We add all the numbers together, and all the variables
(+1x/2)-(+1x/4)+7-((+19/2))=0
We get rid of parentheses
1x/2-1x/4+7-((+19/2))=0
We calculate fractions
(-8x^2)/()+4x/()+7+()/()=0
We add all the numbers together, and all the variables
(-8x^2)/()+4x/()+8=0
We multiply all the terms by the denominator
(-8x^2)+4x+8*()=0
We add all the numbers together, and all the variables
(-8x^2)+4x=0
We get rid of parentheses
-8x^2+4x=0
a = -8; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-8)·0
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-8}=\frac{-8}{-16} =1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-8}=\frac{0}{-16} =0 $
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