A(r)=-2/3r-4

Solution for A(r)=-2/3r-4 equation:

(A)=-2/3A-4

We move all terms to the left:

(A)-(-2/3A-4)=0


Domain of the equation: 3A-4)!=0

A∈R


We get rid of parentheses

A+2/3A+4=0

We multiply all the terms by the denominator


A*3A+4*3A+2=0

Wy multiply elements

3A^2+12A+2=0

a = 3; b = 12; c = +2;
Δ = b2-4ac
Δ = 122-4·3·2
Δ = 120
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{30}}{2*3}=\frac{-12-2\sqrt{30}}{6}
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{30}}{2*3}=\frac{-12+2\sqrt{30}}{6}
$