2(4^2x+1)+40=8

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Solution for 2(4^2x+1)+40=8 equation: 



2(4^2x+1)+40=8

We move all terms to the left:

2(4^2x+1)+40-(8)=0

We add all the numbers together, and all the variables

2(4^2x+1)+32=0

We multiply parentheses

8x^2+2+32=0

We add all the numbers together, and all the variables

8x^2+34=0

a = 8; b = 0; c = +34;
Δ = b2-4ac
Δ = 02-4·8·34
Δ = -1088
Delta is less than zero, so there is no solution for the equation

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