r-(5/4)+(1/2)r=(13/4)

Solution for r-(5/4)+(1/2)r=(13/4) equation:

r-(5/4)+(1/2)r=(13/4)

We move all terms to the left:

r-(5/4)+(1/2)r-((13/4))=0


Domain of the equation: 2)r!=0

r!=0/1

r!=0

r∈R


We add all the numbers together, and all the variables

r+(+1/2)r-(+5/4)-((+13/4))=0

We multiply parentheses

r^2+r-(+5/4)-((+13/4))=0

We get rid of parentheses

r^2+r-5/4-((+13/4))=0

We calculate fractions


r^2+r=0

a = 1; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·1·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*1}=\frac{-2}{2}
=-1
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*1}=\frac{0}{2}
=0
$