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12-(1/5)x=2x+1
We move all terms to the left:
12-(1/5)x-(2x+1)=0
Domain of the equation: 5)x!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
-(+1/5)x-(2x+1)+12=0
We multiply parentheses
-x^2-(2x+1)+12=0
We get rid of parentheses
-x^2-2x-1+12=0
We add all the numbers together, and all the variables
-1x^2-2x+11=0
a = -1; b = -2; c = +11;
Δ = b2-4ac
Δ = -22-4·(-1)·11
Δ = 48
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4\sqrt{3}}{2*-1}=\frac{2-4\sqrt{3}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4\sqrt{3}}{2*-1}=\frac{2+4\sqrt{3}}{-2} $
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