A quadratic equation.... help please?

Given:
Perimeter of rectangle = 50 cm.
Area of the rectangle = 154 cm
Find L = length : W = width

Area of rectangle = L x W
L x W = 154 ====> eq 1
Perimeter of the rectangle = L + W + L + W
Perimeter of the rectangle = 2L + 2 W
2L + 2W = 50 cm
L + W = 25 cm === > eq 2

From eq 2
L + W = 25 cm === > eq 2
L = 25 - W

Substitute L to eq 1
L x W = 154 ====> eq 1

(25 - W ) x W =154
25W - (W)^2 =154

(W)^2 - 25W + 154 =0

(W +11) (W-14) =0

W = -11 & W= 14

Using W =14 and Substitute to eq 2
L + W = 25 cm === > eq 2
L + (14) = 25
L = 25 -14
L = 11

=============================================
Ans:
W = 14 then L = 11
=============================================

p= 2*(w + l) = 50
a = w*l = 154

2w +2l = 50

w*l = 154 => l = 154/w

=> 2w +2*154/w = 50

w +154/w = 25

w^2/w +154/w = 25
w^2 +154 = 25w
w^2 - 25w +154 = 0

use quadratic formula to solve for w

or complete the square

w^2 - 25w = -154
w^2 - 25w + 156.25 = -154 + 156.25 //add (half 25)^2 to each side

w^2 - 25w + 156.25 = 2.25
(w - 12.5)^2 = 2.25

w - 12.5 = square root(2.25)
w - 12.5 = +1.5, -1.5
w = +1.5 + 12.5, -1.5 + 12.5
w = 14, 11

using: w*l = 154

l = 154/w
l = 154/14 = 11
l = 154/11 = 14

so rectangle is 11 by 14

x = width
y = length

2(x + y) = 50
xy = 154

2(x + y) = 50
x + y = 25
y = 25 - x

xy = 154
x(25 - x) = 154
25x - x^2 = 154
x^2 - 25x + 154 = 0
(x - 14)(x - 11) = 0

x - 14 = 0
x = 14

x - 11 = 0
x = 11

2(x + y) = 50
14 + y = 25
y = 25 - 14
y = 11

2(x + y) = 50
11 + y = 25
y = 25 - 11
y = 14

∴ [x (width) = 14 , y (length) = 11] or [x (width) = 11 , y (length) = 14]

2x + 2y = 50
x y = 154
y = 154/x

2x + 308 / x = 50
2x² + 308 = 50x
2x² - 50x + 308 = 0
x² - 25x + 154
(x - 11)(x - 14) = 0
x = 11 , x = 14

let
p - perimeter
a - area
l - lenght of the rectangle
w - width of the rectangle

given:
p = 50cm
a = 142 cm^2

formula:
p = 2l + 2w
a = lw

substitute the given to the respective formulas
50 = 2l+2w
divide the above equation by GCF 2
25 = l + w
l = 25 - w

substitute l
154 = lw
154 = (25-w)w
distribute
154 = 25w - w^2

tranpose 25w and -w^2 to the right side of the equation
w^2 - 25w + 154 = 0
factor
(w - 14)(w - 11) = 0
w - 14 = 0
w = 14cm
l = 25 - w = 25 - 14 = 11cm

w-11 = 0
w = 11cm
l = 25 - w = 25 - 11 = 14 cm

hope that's okay
thanks

2(L+B) = 50
LB = 154

L = 154/B

So, 2L+2B = 50
So, 2*154/B + 2B = 50
Dividing throughout by 2 gives,
154/B + B = 25
So, (154+B2) = 25
So, B2 - 25B + 154 = 0
So, B2 - 14B - 11B + 154 = 0
So, B(B-14) - 11 (B-14) = 0
Taking common,
(B - 14)(B - 11) = 0
So, B-14 = 0 OR B-11 = 0
So, B = 14 or B = 11

If B = 14, L = 11 (using 2L+2B=50)
If B = 11, L = 14

But, length will always be greater than breadth, so L = 14 and B = 11

say, length and breadth of rectangle are l and b, resectively.
perimeter of rectangle=2(l+b)=50
ie, l+b=25.. ie, b=25-l
area of rectangle=lb=154
ie, l(25-l)=154
ie, 25l-l^2=154
ie, l^2-25l+154=0
ie, l^2-11l-14l+154=0
ie, l(l-11)-14(l-11)=0
ie,(l-14)(l-11)=0
l=11,14
as lb=154, so b=14,11
so, the length and breadth are 14,11 or 11,14.

half perimeter, w + l = 25
area : w*l = 154

2w*l = 308
(w + l)^2 = 25^2 = 625
w^2 + l^2 + 2w*l = 625
w^2 + l^2 = 625 - 308 = 317

w^2 + l^2 - 2w*l = 317 - 308 = 9 = (w - l)^2
w - l = 3
w + l = 25
==>
2w = 28 ==> w = 14, l = 11

Let y be the length of the rectangle.
Let x be the width of the rectangle.

Now, the perimeter is (y + y + x + x), right?

Therefore:

y + y + x + x = 50
2y + 2x = 50
2y = 50 - 2x

y = 25 - x

Now we have y by itself:

We also know that area is xy (length multiplied by width) so we have:

xy = 154

Remember, we established that y = 25 - x before so simply replace the y in the above equation with 25 - x

x(25 - x) = 154
25x - x^2 = 154
x^2 - 25x + 154 = 0

x = 11 or 14

Therefore, the width is 11cm or 14cm.

That means the rectangle has dimensions 11x14 cm.

x+y=25
xy=154
factors of 154 is 2,7,11
therefore the answer is

11,14