PLS HELP!!!

We are asked to find: \(x^2+y^2\), for \(x\) and \(y\) are the roots of the equation.

\(3x^2+4x+12\) has a sum of \(\frac{-4}{3}.\)

\(3x^2+4x+12\) has a product of \(\frac{12}{3}=4.\)

Use the manipulation:\((x+y)^2-2(xy)=x^2+y^2\).

Thus, the answer is \((\frac{-4}{3})^2-2(4)=\frac{16}{9}-\frac{72}{9}=\frac{-56}{9}.\)

By Vieta....

We have  the form   Ax^2 + Bx + C  = 0

Let the roots  be  R₁ and R₂

The sum of these  =  -B/A  =   -4/3  =   R₁ + R₂     (1)

And the product of these =  C/A  =   =  12/3 =  4  = R₁*R₂  which implies that  8 = 2*R₁*R₂

Square  (1)  and we get that

R₁^2  + 2*R₁*R₂  + R₂^2  =  16/9

R₁^2  + 8  + R₂^2  =  16/9

R₁^2  + 72/9  + R₂^2  = 16/9      subtract 72/9  from both sides

R₁^2 + R₂^2  =  16/9 - 72/9  =  -56/9

THX SO MUCH GUYS!!!!!!