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2^x-10/2^x+3=0
Domain of the equation: 2^x!=0We multiply all the terms by the denominator
x!=0/1
x!=0
x∈R
2^x*2^x+3*2^x-10=0
Wy multiply elements
4x^2+6x-10=0
a = 4; b = 6; c = -10;
Δ = b2-4ac
Δ = 62-4·4·(-10)
Δ = 196
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-14}{2*4}=\frac{-20}{8} =-2+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+14}{2*4}=\frac{8}{8} =1 $
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