Find the zeros by using the quadratic formula and tell whether the solutions are real or imaginary. F(x)=x^2+14x+20.
we have that
the quadratic formula is equal to
[tex]x=\frac{-b\pm\sqrt[\square]{b^2-4ac}_{}}{2a}[/tex]where
a=1
b=14
c=20
substitute the given values in the formula
[tex]\begin{gathered} x=\frac{-14\pm\sqrt[\square]{14^2-4(1)(20)}_{}}{2(1)} \\ \\ x=\frac{-14\pm\sqrt[\square]{116}_{}}{2} \\ \\ x1=\frac{-14+\sqrt[\square]{116}_{}}{2} \\ \\ x2=\frac{-14-\sqrt[\square]{116}_{}}{2} \end{gathered}[/tex]x1 and x2 are the zeros of the quadratic equation
The solutions are real numbers
