3x+((5-4x)^(1/2))=2

Solution for 3x+((5-4x)^(1/2))=2 equation:

3x+((5-4x)^(1/2))=2

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


3x*2))+((-4x+1+5)^(-2*2))=0

We add all the numbers together, and all the variables

3x*2))+((-4x+6)^(-4))=0

We add all the numbers together, and all the variables

-4x+3x*2))+((=0

Wy multiply elements

6x^2-4x=0

a = 6; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·6·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*6}=\frac{0}{12}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*6}=\frac{8}{12}
=2/3
$