2^x+4=3/2x+5

Solution for 2^x+4=3/2x+5 equation:

2^x+4=3/2x+5

We move all terms to the left:

2^x+4-(3/2x+5)=0


Domain of the equation: 2x+5)!=0

x∈R


We get rid of parentheses

2^x-3/2x-5+4=0

We multiply all the terms by the denominator


2^x*2x-5*2x+4*2x-3=0

Wy multiply elements

4x^2-10x+8x-3=0

We add all the numbers together, and all the variables

4x^2-2x-3=0

a = 4; b = -2; c = -3;
Δ = b2-4ac
Δ = -22-4·4·(-3)
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{13}}{2*4}=\frac{2-2\sqrt{13}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{13}}{2*4}=\frac{2+2\sqrt{13}}{8}
$