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X+3/4X=3
We move all terms to the left:
X+3/4X-(3)=0
Domain of the equation: 4X!=0We multiply all the terms by the denominator
X!=0/4
X!=0
X∈R
X*4X-3*4X+3=0
Wy multiply elements
4X^2-12X+3=0
a = 4; b = -12; c = +3;
Δ = b2-4ac
Δ = -122-4·4·3
Δ = 96
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{6}}{2*4}=\frac{12-4\sqrt{6}}{8} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{6}}{2*4}=\frac{12+4\sqrt{6}}{8} $
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