2x+(1/2x+5)=10

Solution for 2x+(1/2x+5)=10 equation:

2x+(1/2x+5)=10

We move all terms to the left:

2x+(1/2x+5)-(10)=0


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

x∈R


We get rid of parentheses

2x+1/2x+5-10=0

We multiply all the terms by the denominator


2x*2x+5*2x-10*2x+1=0

Wy multiply elements

4x^2+10x-20x+1=0

We add all the numbers together, and all the variables

4x^2-10x+1=0

a = 4; b = -10; c = +1;
Δ = b2-4ac
Δ = -102-4·4·1
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{21}}{2*4}=\frac{10-2\sqrt{21}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{21}}{2*4}=\frac{10+2\sqrt{21}}{8}
$