Explanation:
"find" ?
the equations are already there.
do you mean solve in the sense of finding the zero points of the equations ?
if so,
the general solution of a quadratic equation is
x = (-b ± sqrt(b² - 4ac))/(2a)
3(3x² + 1) = 12x
9x² + 3 = 12 x
9x² - 12x + 3 = 0
a = 9
b = -12
c = 3
x = (12 ± sqrt(144 - 4×9×3))/18 = (12 ± sqrt(144-108))/18 =
= (12 ± sqrt(36))/18 = (12 ± 6)/18
x1 = (12 + 6) / 18 = 18/18 = 1
x2 = (12 - 6) / 18 = 6/18 = 1/3 = 0.3333333...
9y² - 12y - 14 = 0
same thing, we just use y instead of x.
a = 9
b = -12
c = -14
y = (12 ± sqrt(144 - 4×9×-14))/18 = (12 ± sqrt(144+504)/18 =
= (12 ± sqrt(648))/18 = (12 ± sqrt(324×2))/18 =
= (12 ± 18×sqrt(2))/18
y1 = (12 + 18×sqrt(2))/18 = 12/18 + sqrt(2) = 2/3 + sqrt(2) =
= 0.942809042...
y2 = (12 - 18×sqrt(2))/18 = 12/18 - sqrt(2) = 2/3 - sqrt(2) =
= -0.747546896...