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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problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

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step-by-step explanation: