Answer:
the solutions are:
[tex]x_{1}=-2+\sqrt{26}[/tex]
[tex]x_{2}=-2-\sqrt{26}[/tex]
and the sum of solutions is:
-4
Step-by-step explanation:
the expression is:
[tex]3x(x+4)=66\\[/tex]
doing the multiplication
[tex]3x^2+12x=66\\\\3x^2+12x-66=0[/tex]
dividing everything by 3:
[tex]x^2+4x-22=0[/tex]
we use the quadratic formula to find the solutions:
in this case since we have an equation in the form
[tex]ax^2+bx+c=0[/tex]
where
[tex]a=1\\b=4\\c=-22[/tex]
the quadratic formula is:
[tex]x=\frac{-b+-\sqrt{b^2-4ac} }{2a} \\\\x=\frac{-4+-\sqrt{4^2-4*1*(-22))=} }{2}\\ x=\frac{-4+-\sqrt{16+88} }{2} \\ x=\frac{-4+-\sqrt{104} }{2} \\\\ x=\frac{-4+-2\sqrt{26} }{2} \\\\\\x=-4+-\sqrt{26}[/tex]
from this, using the + sign we find the first solution:
[tex]x_{1}=-2+\sqrt{26}[/tex]
and the second solution:
[tex]x_{2}=-2-\sqrt{26}[/tex]
the sum of the solutions is:
[tex]x_{1}+x_{2}=-2+\sqrt{26}-2-\sqrt{26}\\ x_{1}+x_{2}=-4[/tex]
