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1/5m-4=2m+3
We move all terms to the left:
1/5m-4-(2m+3)=0
Domain of the equation: 5m!=0We get rid of parentheses
m!=0/5
m!=0
m∈R
1/5m-2m-3-4=0
We multiply all the terms by the denominator
-2m*5m-3*5m-4*5m+1=0
Wy multiply elements
-10m^2-15m-20m+1=0
We add all the numbers together, and all the variables
-10m^2-35m+1=0
a = -10; b = -35; c = +1;
Δ = b2-4ac
Δ = -352-4·(-10)·1
Δ = 1265
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:m_{1}=\frac{-b-\sqrt{\Delta}}{2a}m_{2}=\frac{-b+\sqrt{\Delta}}{2a}m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-\sqrt{1265}}{2*-10}=\frac{35-\sqrt{1265}}{-20}m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+\sqrt{1265}}{2*-10}=\frac{35+\sqrt{1265}}{-20}
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