Please help, due soon I thought I finished but I had 2 more left and I did number 7 already.

We move all terms to the left:

r^2+6-(-5r)=0We get rid of parenthesesr^2+5r+6=0a = 1; b = 5; c = +6;Δ = b2-4acΔ = 52-4·1·6Δ = 1The delta value is higher than zero, so the equation has two solutionsWe use following formulas to calculate our solutions:r1=−b−Δ√2ar2=−b+Δ√2aΔ−−√=1√=1r1=−b−Δ√2a=−(5)−12∗1=−62=−3r2=−b+Δ√2a=−(5)+12∗1=−42=−2

Step-by-step explanation: