How do I solve these problems?

Question

1.x^2+8x= -16
2.3x^2=2x+5

Answer 1

x^2 + 8x +16 = 0
(x+4)(x+4)
x= -4
plug it back into the equation to see if that is the right x

3x^2=2x+5
3x^2-2x-5=0
(3x - 5) (x+1)
x = ( 5/3, -1)

Answer 2

The first problem:

x^2 + 8x = -16
Add 16.
x^2 + 8x + 16 = 0
Then factor.
(x + 4)^2 = 0
Therefore x = -4

The 2nd do the same:
3x^2 - 2x + 5 = 0
(3x - 5)(x + 1) = 0
Therefore x = -1, x = 5/3

Answer 3

x^2 + 8x + 16 = 0
(x + 4) (x + 4) = 0
x = -4

3x^2 = 2x + 5

3x^2- 2x -5 = 0
(3x-5)(x+1) = 0

3x -5 = 0 ... x = 5/3
x + 1 = 0... x = -1

x = 5/3 and x = -1

Answer 4

x^2 + 8x = -16 can be rewritten as x^2 + 8x + 16 = 0 and this can be factored as (x + 4)^2 = 0 and thus x = -4 (double root)

Answer 5

1.x^2+8x=-16
=>x^2+8x+16=0
=>(x)^2+2*x*4+(4)^2=0
=>(x+4)^2=0
=>x+4=0
=>x= -4
2.3x^2=2x+5
=>3x^2-2x-5=0
=>3x^2+3x-5x-5=0
=>3x(x+1)-5(x+1)=0
=>(x+1)(3x-5)=0
Hence either x+1=0 or 3x-5=0
If x+1=0,then x= -1
If 3x-5=0,then 3x=5 or x=5/3
Therefore x= -1 or 5/3

Answer 6

In each case, you put all the numbers to one side, in each case equaling zero.
Then you factor them.
In the first case, that factors to (x+4) (x+4).
Each of the factors (identical in this case) is set to equal 0.
x+4 = 0 yields the answer -4.
On to the second equation:
When factored, the factors are (3x-5) (x+1).
When you set each of these factors equal to zero, you get:
3x-5=0
3x=5
x=5/3
And:
x+1=0
x=-1
Done!

Answer 7

x^2+8x= -16
x^2+8x+16=0
(x+4)(x+4)=0
x= -4
or
x= -4

3x^2=2x+5
3x^2-2x-5=0
(3x-5)(x+1)=0
x=5/3
or
x= -1

Answer 8

those are quadratics so they take the form

1. x^2+8x+16=0

2. 3x^2-2x-5=0

the general form for a quadratic is

x=(-b+-SQRT(b^2-ac))/2a

1. -8+-SQRT(8^2-4(1)(16))/2(1)

-8+-SQRT(64-64)/2

x=-4

2. -2+-SQRT(-2^2-4*3*-5)/2*3

-2+-SQRT(4+60)/6

x=1 or x=1.66666666

Answer 9

use the quadratic equation

Answer 10

i'm not sure but i will tell you this your gonna have to be very very very very very very very very very....... very careful