2^x3-4x2-12x-5=1/32

Solution for 2^x3-4x2-12x-5=1/32 equation:

2^x3-4x^2-12x-5=1/32

We move all terms to the left:

2^x3-4x^2-12x-5-(1/32)=0

We add all the numbers together, and all the variables

-4x^2+2^x3-12x-5-(+1/32)=0

We add all the numbers together, and all the variables

-4x^2-12x+2^x3-5-(+1/32)=0

We get rid of parentheses

-4x^2-12x+2^x3-5-1/32=0

We multiply all the terms by the denominator


-4x^2*32-12x*32+2^x3*32-1-5*32=0

We add all the numbers together, and all the variables

-4x^2*32-12x*32+2^x3*32-161=0

Wy multiply elements

-128x^2-384x+64x-161=0

We add all the numbers together, and all the variables

-128x^2-320x-161=0

a = -128; b = -320; c = -161;
Δ = b2-4ac
Δ = -3202-4·(-128)·(-161)
Δ = 19968
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:

$\sqrt{\Delta}=\sqrt{19968}=\sqrt{256*78}=\sqrt{256}*\sqrt{78}=16\sqrt{78}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-320)-16\sqrt{78}}{2*-128}=\frac{320-16\sqrt{78}}{-256}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-320)+16\sqrt{78}}{2*-128}=\frac{320+16\sqrt{78}}{-256}
$