R^2+2r-28=-3r-3-r^2

Solution for R^2+2r-28=-3r-3-r^2 equation:

^2+2R-28=-3R-3-R^2

We move all terms to the left:

^2+2R-28-(-3R-3-R^2)=0

We add all the numbers together, and all the variables

-(-3R-3-R^2)+2R=0

We get rid of parentheses

R^2+3R+2R+3=0

We add all the numbers together, and all the variables

R^2+5R+3=0

a = 1; b = 5; c = +3;
Δ = b2-4ac
Δ = 52-4·1·3
Δ = 13
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{-(5)-\sqrt{13}}{2*1}=\frac{-5-\sqrt{13}}{2}
$
$R_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{13}}{2*1}=\frac{-5+\sqrt{13}}{2}
$