We are given the following two equations
[tex]\begin{gathered} y=(x+4)^2-1\qquad eq.1_{} \\ y=-3x-13\qquad eq.2 \end{gathered}[/tex]Since both equations are equal to y then we can equate them as
[tex](x+4)^2-1=-3x-13[/tex]Now let us simplify the equation
[tex]\begin{gathered} (x+4)^2-1=-3x-13 \\ x^2+2(x)(4)+4^2-1=-3x-13 \\ x^2+8x+16-1=-3x-13 \\ x^2+8x+3x+16-1+13=0 \\ x^2+11x+28=0 \end{gathered}[/tex]As you can see, we are left with a quadratic equation.
The standard form of a quadratic equation is given by
[tex]ax^2+bx+c=0[/tex]Comparing the equation with the standard form, the coefficients are
a = 1
b = 11
c = 28
Now recall that the quadratic formula is given by
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]Substitute the values into the above quadratic formula
[tex]x=\frac{-11\pm\sqrt[]{(11)^2-4(1)(28)}}{2(1)}=\frac{-11\pm\sqrt[]{121^{}-112}}{2}=\frac{-11\pm\sqrt[]{9}}{2}=\frac{-11\pm3}{2}[/tex]So the two possible solutions are
[tex]\begin{gathered} x_1=\frac{-11+3}{2},\: x_2=\frac{-11-3}{2} \\ x_1=\frac{-8}{2},\: x_2=\frac{-14}{2} \\ x_1=-4,\: x_2=-7 \end{gathered}[/tex]Therefore, the solutions of the equations are
x = -4 and x = -7