-r2+6r-5=-3r2+3r-2

Solution for -r2+6r-5=-3r2+3r-2 equation:

-r2+6r-5=-3r^2+3r-2

We move all terms to the left:

-r2+6r-5-(-3r^2+3r-2)=0

We add all the numbers together, and all the variables

-1r^2-(-3r^2+3r-2)+6r-5=0

We get rid of parentheses

-1r^2+3r^2-3r+6r+2-5=0

We add all the numbers together, and all the variables

2r^2+3r-3=0

a = 2; b = 3; c = -3;
Δ = b2-4ac
Δ = 32-4·2·(-3)
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{33}}{2*2}=\frac{-3-\sqrt{33}}{4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{33}}{2*2}=\frac{-3+\sqrt{33}}{4}
$