Answer:
Option A. [tex]-10[/tex]
Step-by-step explanation:
we know that
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2} +10x+24=0[/tex]
so
[tex]a=1\\b=10\\c=24[/tex]
substitute in the formula
[tex]x=\frac{-10(+/-)\sqrt{10^{2}-4(1)(24)}} {2(1)}[/tex]
[tex]x=\frac{-10(+/-)\sqrt{4}} {2}[/tex]
[tex]x=\frac{-10(+/-)2} {2}[/tex]
[tex]x=\frac{-10(+)2} {2}=-4[/tex]
[tex]x=\frac{-10(-)2} {2}=-6[/tex]
The roots are -4 and -6
Find the sum
[tex]-4-6=-10[/tex]
