Answer:
The roots of the equation : [tex]x^2+14x+50=0[/tex] are [tex]\mathbf{x=-7+i \ or \ x=-7-i}[/tex]
Step-by-step explanation:
We need to find roots of the equation : [tex]x^2+14x+50=0[/tex]
We can solve the equation using quadratic formula: [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
From equation we get: a=1, b=14 and c=50
Putting values and finding roots
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-14\pm\sqrt{(14)^2-4(1)(50)}}{2(1)}\\x=\frac{-14\pm\sqrt{196-200}}{2(1)}\\x=\frac{-14\pm\sqrt{-4}}{2}\\We \ know \ \sqrt{-1}=i \\x=\frac{-14\pm2i}{2}\\x=\frac{-14+2i}{2} \ or \ x=\frac{-14-2i}{2}\\x=\frac{2(-7+i)}{2} \ or \ x=\frac{2(-7-i)}{2}\\x=-7+i \ or \ x=-7-i[/tex]
So, the roots of the equation : [tex]x^2+14x+50=0[/tex] are [tex]\mathbf{x=-7+i \ or \ x=-7-i}[/tex]
