Answer:
The solutions are [tex]-2[/tex] and [tex]-8[/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+16=0[/tex]
so
[tex]a=1\\b=10\\c=16[/tex]
substitute in the formula
[tex]x=\frac{-10(+/-)\sqrt{10^{2}-4(1)(16)}} {2(1)}[/tex]
[tex]x=\frac{-10(+/-)\sqrt{36}}{2}[/tex]
[tex]x=\frac{-10(+/-)6}{2}[/tex]
[tex]x=\frac{-10(+)6}{2}=-2[/tex]
[tex]x=\frac{-10(-)6}{2}=-8[/tex]
The solutions are [tex]-2[/tex] and [tex]-8[/tex]
