how do you use the quadratic formula to solve 3x^2+4x=6

Question

anonymous · Answer

move the 6 to the other side so you have  $$3x^2+4x-6=0$$ then you have your A,B, and C

anonymous · Answer

how do you finish the problem

anonymous · Answer

$$ax^2 +bx +c$$

anonymous · Answer

"-b +/- sqrt(b^2-4ac)" all divided by "2a"

anonymous · Answer

Set it equal to 0 
3x^2 +4x -6 =0
Now use the quadratic formal to solve for x. 
x= (-4+-sqroot[4^2-4(3)(-6)]/6 
=(-4+-sqroot[16+72]/6
=(-4+-sqroot[88]/6:
x=(-4+sqroot{88]/6 or x=(-4-sqroot[88]/6
simplify radicals. 
sqroot[88] =sqroot[2^3x11] = 2sqroot[2x11]= 2sqroot[11]:
x=(-4+ 2sqr[22])/6 or x=(-4-2sqr[22])/6
x=((1/3)sqrt[[22]-2]) or x=((1/3) (-2-sqrt[22]))