Question

3x^2+x+8<12
please solve usiing quadraic inequalities

Answer 1

start with
\[3x^2+x-4<0\] then factor as
\[(x-1)(3x+4)<0\] since this is a parabola that opens up, it is negative between the zeros, which are \(-\frac{4}{3}\) and \(1\)

Answer 2

THANKS:)

Answer 3

yw