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1/5(25x-75)-4=1/2(2x+54)
We move all terms to the left:
1/5(25x-75)-4-(1/2(2x+54))=0
Domain of the equation: 5(25x-75)!=0
x∈R
Domain of the equation: 2(2x+54))!=0We calculate fractions
x∈R
(2x2/(5(25x-75)*2(2x+54)))+(-5x2/(5(25x-75)*2(2x+54)))-4=0
We calculate terms in parentheses: +(2x2/(5(25x-75)*2(2x+54))), so:
2x2/(5(25x-75)*2(2x+54))
We multiply all the terms by the denominator
2x2
We add all the numbers together, and all the variables
2x^2
Back to the equation:
+(2x^2)
We calculate terms in parentheses: +(-5x2/(5(25x-75)*2(2x+54))), so:We add all the numbers together, and all the variables
-5x2/(5(25x-75)*2(2x+54))
We multiply all the terms by the denominator
-5x2
We add all the numbers together, and all the variables
-5x^2
Back to the equation:
+(-5x^2)
2x^2+(-5x^2)-4=0
We get rid of parentheses
2x^2-5x^2-4=0
We add all the numbers together, and all the variables
-3x^2-4=0
a = -3; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·(-3)·(-4)
Δ = -48
Delta is less than zero, so there is no solution for the equation
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