-2r^2+5r-22=-4r^2+6r+6

Solution for -2r^2+5r-22=-4r^2+6r+6 equation:

-2r^2+5r-22=-4r^2+6r+6

We move all terms to the left:

-2r^2+5r-22-(-4r^2+6r+6)=0

We get rid of parentheses

-2r^2+4r^2-6r+5r-6-22=0

We add all the numbers together, and all the variables

2r^2-1r-28=0

a = 2; b = -1; c = -28;
Δ = b2-4ac
Δ = -12-4·2·(-28)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-15}{2*2}=\frac{-14}{4}
=-3+1/2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+15}{2*2}=\frac{16}{4}
=4
$