Answer:
The given equation has no solution.
Step-by-step explanation:
Given the equation [tex]x^2+14x+112=0[/tex]
we have to find the solution of the above equation.
The roots of quadratic equation [tex]ax^2+bx+c=0[/tex] is
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Equation : [tex]x^2+14x+112=0[/tex]
The roots are
[tex]x=\frac{-14\pm \sqrt{14^2-4(1)(112)}}{2}[/tex]
[tex]x=\frac{-14\pm \sqrt{196-448}}{2}=\frac{-14\pm \sqrt{-252}}{2}[/tex]
Since, determinant under the root is negative
⇒ The given equation has no real solution.
Last option is correct.
