4r-3=3(3r+4)4r-3=3(3r+4)

Solution for 4r-3=3(3r+4)4r-3=3(3r+4) equation:

4r-3=3(3r+4)4r-3=3(3r+4)

We move all terms to the left:

4r-3-(3(3r+4)4r-3)=0


We calculate terms in parentheses: -(3(3r+4)4r-3), so:

3(3r+4)4r-3

We multiply parentheses

36r^2+48r-3

Back to the equation:

-(36r^2+48r-3)


We get rid of parentheses

-36r^2+4r-48r+3-3=0

We add all the numbers together, and all the variables

-36r^2-44r=0

a = -36; b = -44; c = 0;
Δ = b2-4ac
Δ = -442-4·(-36)·0
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-44)-44}{2*-36}=\frac{0}{-72}
=0
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-44)+44}{2*-36}=\frac{88}{-72}
=-1+2/9
$