1/2r+4=r+10

Solution for 1/2r+4=r+10 equation:

1/2r+4=r+10

We move all terms to the left:

1/2r+4-(r+10)=0


Domain of the equation: 2r!=0

r!=0/2

r!=0

r∈R


We get rid of parentheses

1/2r-r-10+4=0

We multiply all the terms by the denominator


-r*2r-10*2r+4*2r+1=0

Wy multiply elements

-2r^2-20r+8r+1=0

We add all the numbers together, and all the variables

-2r^2-12r+1=0

a = -2; b = -12; c = +1;
Δ = b2-4ac
Δ = -122-4·(-2)·1
Δ = 152
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{152}=\sqrt{4*38}=\sqrt{4}*\sqrt{38}=2\sqrt{38}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{38}}{2*-2}=\frac{12-2\sqrt{38}}{-4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{38}}{2*-2}=\frac{12+2\sqrt{38}}{-4}
$