Question

How do you solve the equation `3x^2+6x-1=8`?

Answer 1

You want an equation like `ax^2+bx+c=0`, so we subtract 8 from both sides. We now get `3x^2+6x-9 = 0`, so we find `a=3, b=6` and `c=-9`.
We use the quadratic formula to solve this equation:
`D=b^2-4ac = 6^2-4*3*-9 = 36+108=144.`
We know that `x=(-b+-sqrt(D))/(2a) = (-6+-sqrt(144))/(2*3) = (-6+-12)/6= -1+-2`, so `x=1 vee x=-3`.