r^2+3r-2=0 Solve the equation using the quadratic formula?

two different answers, try for 3

r²+3r-2=0
quadratic equation:
r = [-b ±√(b²-4ac)] / 2a
r = [-3 ±√(9+8)] / 2
r = [-3 ±√17] / 2
r = [-3 ±4.123] / 2
r = -3.562, +0.562

Using the quadratic formula, note that in your given function, the following are applicable:

a = 1

b = 3

c = -1

Simply substitute the appropriate values in the quadratic formula and your roots will be

r = 0.56 and r = -3.56.

r= [-3 +-sqroot(3^2-4*1*(-2))]/2
r=[-3+-sqroot(17) ]/2

2solutions:
r1=0.56155
r2=-3.56155

r^2 + 3r - 2 = 0
r = [-b ±√(b^2 - 4ac)]/2a

a = 1
b = 3
c = -2

r = [-3 ±√(9 + 8)]/2
r = [-3 ±√17]/2
r = [-3 ±4.12]/2 (approx.)

r = [-3 + 4.12]/2
r = 1.12/2
r = 0.56

r = [-3 - 4.12]/2
r = -7.12/2
r = -3.56

∴ r = 0.56 , -3.56 (approx.)

r = [ - 3 ± √ (9 + 8 ) ] / 2
r = [ - 3 ± √17 ] / 2
r = 0.56 , r = - 3.56

r²+3r-2=0

quadratic formula:
x= [-b±√(b²-4ac)]/2a
x=[-3±√(3²-4*1*-2)]/2*1
x=[-3±√(9-8)]/2
x=[-3±√(1)]/2
x=[-3±1]/2
x=[-3 - 1]/2 and x=[-3 + 1]/2
x=[-2]/2 and x=[-4]/2
x=-1 and x=-2
{-1,-2}

Quad. formula.

-b + or - sqrt. of b^2 - 4ac all over 2a

Equation r^2 + 3r - 2 = 0

-3 + or - sqrt. of 3^2 - 4(1)(-2) all over 2(1)

-3 + or - sqrt. of 9 + 8 all over 2

(-3 + or - sqrt. of 17)/ 2

r =
(-3 + sqrt. of 17)/ 2 and (-3 - sqrt. of 17)/ 2

Δ = 3^2 - 4*1*(-2) = 9 + 8 = 17

r = [-3 + sqrt(Δ)]/2 or r = [-3 - sqrt(Δ)]/2

r = r = [-3 + sqrt(17)]/2 or r = [-3 - sqrt(17)]/2

quadratic equation : (-b +/- √(b^2 - 4ac)) / 2a

(-3 + √(9-8)) / 2 = -1
(-3 - √(9-8)) / 2 = -2