-(2^x)=3/4x-4

Solution for -(2^x)=3/4x-4 equation:

-(2^x)=3/4x-4

We move all terms to the left:

-(2^x)-(3/4x-4)=0


Domain of the equation: 4x-4)!=0

x∈R


We get rid of parentheses

-2^x-3/4x+4=0

We multiply all the terms by the denominator


-2^x*4x+4*4x-3=0

Wy multiply elements

-8x^2+16x-3=0

a = -8; b = 16; c = -3;
Δ = b2-4ac
Δ = 162-4·(-8)·(-3)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{10}}{2*-8}=\frac{-16-4\sqrt{10}}{-16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{10}}{2*-8}=\frac{-16+4\sqrt{10}}{-16}
$