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2/5x+4=1+x
We move all terms to the left:
2/5x+4-(1+x)=0
Domain of the equation: 5x!=0We add all the numbers together, and all the variables
x!=0/5
x!=0
x∈R
2/5x-(x+1)+4=0
We get rid of parentheses
2/5x-x-1+4=0
We multiply all the terms by the denominator
-x*5x-1*5x+4*5x+2=0
Wy multiply elements
-5x^2-5x+20x+2=0
We add all the numbers together, and all the variables
-5x^2+15x+2=0
a = -5; b = 15; c = +2;
Δ = b2-4ac
Δ = 152-4·(-5)·2
Δ = 265
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{-(15)-\sqrt{265}}{2*-5}=\frac{-15-\sqrt{265}}{-10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{265}}{2*-5}=\frac{-15+\sqrt{265}}{-10} $
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