Yoshi is six times as old as his granddaughter jana. in 3 years, yoshi will be three times as old as the square of jana's age 7 years ago. how old is yoshi now?

Y=6j
in 3 years (y+3) seven years ago (j-7)
y+3=3((j-7)²)
subsitute 6j for y

6j+3=3((j-7)²)
expand
6j+3=3(j²-14j+49)
divide both sides by 3
2j+1=j²-14j+49
minus 2j+1 both sides
0=j²-16j+48
factor
what 2 numbers multiply to get 48 and add to get -16?
-12 and -4
0=(j-12)(j-4)
set to zero
j-12=0
j=12
j-4=0
j=4
we exclude that because we have 'jana's age 7 years ago' she can't be negative age

so jana=12 years old
then yosi=12 itmes 6=72

yoshi is 72 now

Okay, this is a lot of information, but we'll take it one step at a time. Let's let "x" be Yoshi's age and "y" be Jana's age. We know the following:

X = 6y
X = 3 (√y - 7)

Since x both equals the first equation and the second equation, we can set the equations to themselves:

6y = 3 √(y - 7)

Begin simplifying by dividing three from both sides:

(6y)÷3 = √(y - 7)

Square both sides:

(2y)² = y - 7

If you look at this, it's beginning to look like a quadratic. In that case, move everything to one side, and set it to zero:

(2y)² - y + 7 = 0

Use the quadratic formula to solve. You should get two answers, and only one of them will be realistic. If you need a reminder, the quadratic formula is below, and 2 is "a", -1 is "b", and 7 is "c".

y = -b (+-) √(b² - 4ac)

2a

9/12 is the answer because 3/4 times 3 each is 9/12