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41/2x+3=3x+12-x
We move all terms to the left:
41/2x+3-(3x+12-x)=0
Domain of the equation: 2x!=0We add all the numbers together, and all the variables
x!=0/2
x!=0
x∈R
41/2x-(2x+12)+3=0
We get rid of parentheses
41/2x-2x-12+3=0
We multiply all the terms by the denominator
-2x*2x-12*2x+3*2x+41=0
Wy multiply elements
-4x^2-24x+6x+41=0
We add all the numbers together, and all the variables
-4x^2-18x+41=0
a = -4; b = -18; c = +41;
Δ = b2-4ac
Δ = -182-4·(-4)·41
Δ = 980
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{980}=\sqrt{196*5}=\sqrt{196}*\sqrt{5}=14\sqrt{5}x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-14\sqrt{5}}{2*-4}=\frac{18-14\sqrt{5}}{-8}x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+14\sqrt{5}}{2*-4}=\frac{18+14\sqrt{5}}{-8}
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