(3r^2-3)/3=4r^2-6

Solution for (3r^2-3)/3=4r^2-6 equation:

(3r^2-3)/3=4r^2-6

We move all terms to the left:

(3r^2-3)/3-(4r^2-6)=0

We get rid of parentheses

-4r^2+(3r^2-3)/3+6=0

We multiply all the terms by the denominator


-4r^2*3+(3r^2-3)+6*3=0

We add all the numbers together, and all the variables

-4r^2*3+(3r^2-3)+18=0

Wy multiply elements

-12r^2+(3r^2-3)+18=0

We get rid of parentheses

-12r^2+3r^2-3+18=0

We add all the numbers together, and all the variables

-9r^2+15=0

a = -9; b = 0; c = +15;
Δ = b2-4ac
Δ = 02-4·(-9)·15
Δ = 540
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:

$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{15}}{2*-9}=\frac{0-6\sqrt{15}}{-18}
=-\frac{6\sqrt{15}}{-18}
=-\frac{\sqrt{15}}{-3}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{15}}{2*-9}=\frac{0+6\sqrt{15}}{-18}
=\frac{6\sqrt{15}}{-18}
=\frac{\sqrt{15}}{-3}
$