Can someone please help me in Algebra 2 problems

I need Help! I have an exam tomorrow & i'm completing a study guide but i dont know these problems, i dont remember ._.
Directions: Solve the equation.
1.) x^2 + 12x+ 36 +9



Directions: What are the real or imaginary solutions of the polynomial equation?
1.) x^3 - 125 = 0


Please help & explain. Step by Step !(:

x^2 + 12x+ 36 +9 is not an equation (there is no equal sign),

assuming you meant
x^2 + 12x+ 36 = 9

we can write it as
(x+6)^2=3^2

note that square root has two solutions because
(-3)^2=9
and
(3)^2=9


so solutions are

x+6= +3
x+6= -3

or
x=-3
x=-9

x^3 - 125 = 0
x^3 = 125
x^3 = 5^3

one solution is x=5
now, note that in case of power n, there are n solutions. they lie on circle of radius r and 360deg is divided into n segments. here radius is 5 as we have seen from obvious solution x1=5. the other two solutions are
x2=5, angle 120deg
x3=5, angle 240deg
real part is 5*cos(120)=5*cos(240)= -5/2
imaginary part is +/-5*sin(120)=-/+sin(240)= +/- sqrt(3)*5/2

the other way (one of several ways) is to express equation as factored polynomial
(x-x1)(x-x2)(x-x3)=0
we already know that x1=5, so before we can find x2 and x3, we need to get rid of solution we know. we do that by dividing x^3-125 with (x-x1) or in this case (x-5). result is x^2+5x+25 which can be used in quadratic equation to get complex roots:

x2=(5/2)(-1-i*sqrt(3))
x3=(5/2)(-1+i*sqrt(3))

1.) x^2+12x+36+9 = X^2+12*X+45
Standard form: a(X-h)²+k = ( X² +12X +36) -36 +45 = (X +6)² +9
X = -6 ±√( -9) = -6+3i, or -6-3i

( x -5)( x^2 +5x +25)
real: x=5
complex roots: -2.5± 4.330127019i