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1/2^x+3x+8=0
Domain of the equation: 2^x!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
3x+1/2^x+8=0
We multiply all the terms by the denominator
3x*2^x+8*2^x+1=0
Wy multiply elements
6x^2+16x+1=0
a = 6; b = 16; c = +1;
Δ = b2-4ac
Δ = 162-4·6·1
Δ = 232
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{58}}{2*6}=\frac{-16-2\sqrt{58}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{58}}{2*6}=\frac{-16+2\sqrt{58}}{12} $
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