3r²-6-3r=0 i have to use quadratic formula but i dont get how to imput it

r = 2

r = -1

Step-by-step explanation:

So first, we need to get this is order of: ax^2 + bx + c

3r^2 - 6 - 3r

3^2 - 3r - 6

the formula is: -b ± √b^2 -4ac/2a

"ax^2 + bx + c" = "3^2 - 3r - 6"

a = 3

b = -3

c = -6

-b ± √b^2 -4ac/2a

-(-3) ± √(-3)^2 -4(3)(-6)/2(3)

3 ± √9 + (-12)(-6)/6

3 ± √9 + (72)/6

3 ± √9 + 72/6

3 ± √81/6

3 ± 9/6

3 - 9/6

-6/6

-1 = r

3 + 9/6

12/6

2 = r

Formula down below --

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hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)