Lighter, milli is less then centi
L=2+W
A=LW
A=(2+W)W
20=2W+W^2
0=W^2+2W-20
use quadratic
fr
aw^2+bw+C=0
w=(-b+/-sqrt(b^2-4ac))/2a
a=1
b=2
c=-20
w=(-2+/-sqrt(2^2-4(1)(-20)))/2(1)
w=(-2+/-sqrt(4+80))/2
w=(-2+/-sqrt(84))/2
w=(-2+/-2sqrt(21))/2
w=1+/-sqrt(21)
aprox
w=-3.58 or 5.58
cannot have negative width
w=5.58
round
w=5.6
sub
l=2+w
l=2+5.6
l=7.6
legnth=7.6m
width=5.6m
Answer:
Step-by-step explanation:
Lack of information
You could simplify this work by factoring "3" out of all four terms, as follows:
3(x^2 + 2x - 3) =3(0) = 0
Hold the 3 for later re-insertion. Focus on "completing the square" of x^2 + 2x - 3.
1. Take the coefficient (2) of x and halve it: 2 divided by 2 is 1
2. Square this result: 1^2 = 1
3. Add this result (1) to x^2 + 2x, holding the "-3" for later:
x^2 +2x
4 Subtract (1) from x^2 + 2x + 1: x^2 + 2x + 1 -3 -1 = 0,
or x^2 + 2x + 1 - 4 = 0
5. Simplify, remembering that x^2 + 2x + 1 is a perfect square:
(x+1)^2 - 4 = 0
We have "completed the square." We can stop here. or, we could solve for x: one way would be to factor the left side:
[(x+1)-2][(x+1)+2]=0 The solutions would then be:
x+1-2=0=> x-1=0, or x=1, and
x+1 +2 = 0 => x+3=0, or x=-3. (you were not asked to do this).
