1/4^y-3=5-2^y

Solution for 1/4^y-3=5-2^y equation:

1/4^y-3=5-2^y

We move all terms to the left:

1/4^y-3-(5-2^y)=0


Domain of the equation: 4^y!=0

y!=0/1

y!=0

y∈R


We get rid of parentheses

1/4^y+2^y-5-3=0

We multiply all the terms by the denominator


2^y*4^y-5*4^y-3*4^y+1=0

Wy multiply elements

8y^2-20y-12y+1=0

We add all the numbers together, and all the variables

8y^2-32y+1=0

a = 8; b = -32; c = +1;
Δ = b2-4ac
Δ = -322-4·8·1
Δ = 992
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{992}=\sqrt{16*62}=\sqrt{16}*\sqrt{62}=4\sqrt{62}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{62}}{2*8}=\frac{32-4\sqrt{62}}{16}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{62}}{2*8}=\frac{32+4\sqrt{62}}{16}
$