select the approximate values of x that are solutions to f(x)=0, where f(x)=4^2+6x+3

Question

i need help please

Answer 1

If this is meant to be the function f(x)=4x^2+4x+3=0, there are no real solutions, because use of the quadratic formula gives rise to the square root of a negative number (36-48=-12). The minimum point of f(x) is (-0.75,0.75).

Perhaps the first term, 4x^2, should have been negative? If so, then the square root would be of 36+48=84. An approximation to the square root would be square root of 81, which is 9. The approx roots, or zeroes, are x=15/8=1.9 or -3/8=-0.4.