Answer:
The integers are -13 and -12
Step-by-step explanation:
Let
x ----> the first consecutive negative integer
x+1 ----> the second consecutive negative integer
we know that
[tex]x(x+1)=156[/tex]
Apply the distributive property
[tex]x^2+x=156\\x^2+x-156=0[/tex]
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+x-156=0[/tex]
so
[tex]a=1\\b=1\\c=-156[/tex]
substitute in the formula
[tex]x=\frac{-1(+/-)\sqrt{1^{2}-4(1)(-156)}} {2(1)}[/tex]
[tex]x=\frac{-1(+/-)\sqrt{625}} {2}[/tex]
[tex]x=\frac{-1(+/-)25} {2}[/tex]
[tex]x_1=\frac{-1(+)25} {2}=12[/tex] ----> the solution cannot be a positive number
[tex]x_1=\frac{-1(-)25} {2}=-13[/tex]
therefore
[tex]x=-13\\x+1=-12[/tex]
