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4+a-12+3/4a=8
We move all terms to the left:
4+a-12+3/4a-(8)=0
Domain of the equation: 4a!=0We add all the numbers together, and all the variables
a!=0/4
a!=0
a∈R
a+3/4a-16=0
We multiply all the terms by the denominator
a*4a-16*4a+3=0
Wy multiply elements
4a^2-64a+3=0
a = 4; b = -64; c = +3;
Δ = b2-4ac
Δ = -642-4·4·3
Δ = 4048
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4048}=\sqrt{16*253}=\sqrt{16}*\sqrt{253}=4\sqrt{253}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-4\sqrt{253}}{2*4}=\frac{64-4\sqrt{253}}{8} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+4\sqrt{253}}{2*4}=\frac{64+4\sqrt{253}}{8} $
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