Can someone please help me solve for R? The question is r^2+3r=6, thank you.?

Answer 1

r^2+3r-6=0
here a=1.b=3 and c= -6
r={-3+-sqrt(9-(4*-6)}/2
=(-3+-sqrt33)/2

Answer 2

First get the equation equal to 0, si
r^2+3r-6=0
Now use the quadratic formula:
r = [-b +/- sqrt(b^2-4ac)]/2a where a, b, and c are the coefficients from tha above equation.
so r = [-3 +/- sqrt(9-4(1)(-6))]/2
so r = [-3 +/- sqrt(33)]/2
so there are two answers (as always)
r = [-3 + sqrt(33)]/2
and
r = [-3 - sqrt(33)]/2

Answer 3

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

So, use formula, [-b + or - squareroot of ( b^2 - 4ac)]/ 2a
a = 1, b = 3, c = -6

r = [ - 3 + squareroot of ( 3^2 - (4*1*(-6))] / (2*1)
~ 1.37

OR

r = [ - 3 - squareroot of ( 3^2 - (4*1*(-6))] / (2*1)
~ -4.37

Answer 4

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

Solve using quadratic formula

r = (-b ± √(b² - 4ac))/2a

r = (-3 ± √(9 + 24))/2

r = (-3 ± √(33))/2

r = -4.37, r = 1.37
.