Solve 3x2 + 6x = 8. a. 1 plus or minus 2 square root 33
b. the quantity of negative 3 plus or minus I square root 15 all over 3
c. the quantity of negative 3 plus or minus square root 33 all over 3
d. 1 plus or minus 2 i square root 15

The quantity of negative 1 plus or minus square root 33 all over 3

Step-by-step explanation:

3x2+6x=8

3x^2+6x-8=0

Quadratic formula

x=-b +or-√b^2-4ac/2a

a=3

b=6

c=-8

x=-b +or-√b^2-4ac/2a

=-6 +or- √(6)^2 - 4*3*-8/2*3

=-6 +or- √36-(-96)/6

=-6 +or-√36+96/132

=-6 +or- √132/6

=-6 +or- 2√33/6

=-6/6 +or- 2√33/6

= -1 +or- √33/3

The quantity of negative 1 plus or minus square root 33 all over 3

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt

because that’s us why