(3/2*r)+4=10

Solution for (3/2*r)+4=10 equation:

(3/2*r)+4=10

We move all terms to the left:

(3/2*r)+4-(10)=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

(+3/2*r)+4-10=0

We add all the numbers together, and all the variables

(+3/2*r)-6=0

We get rid of parentheses

3/2*r-6=0

We multiply all the terms by the denominator


-6*2*r+3=0

Wy multiply elements

-12r*r+3=0

Wy multiply elements

-12r^2+3=0

a = -12; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-12)·3
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*-12}=\frac{-12}{-24}
=1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*-12}=\frac{12}{-24}
=-1/2
$