1.) solve 2x^2-12x+20=0 using the quadratic solution set? what is the solution set? a.) {3+i, 3-i}b.) {3+2i, 3-i}c.) {2i, -3i}d.) {-3+i, -3-i}2.) lucy divided the two polynomials shown. 3x+2 sqrt 3x^2-x-2a.)x+1b.)x-2+3/3x+2c.) x-1d.) x-2+-3/3x+2

Please, share just ONE problem at a time.  Thanks.

Solve 2x^2-12x+20=0:
Simplify this by dividing each term by 2:  x^2 - 6x + 10 = 0

Identify a, b and c:  a=1, b=-6 and c=10.  Then b^2=36.

Write out the solutions using the quadratic formula:

        6 plus or minus sqrt(36-40)
x =
                            2

sqrt(36-40) = sqrt(-4) = plus or minus i2

Then:
 
         6 plus or minus i2
x = (answer)
                  2

Best answer that i found on yahoo ok   set :   x= length of 2nd piece in cm  x- 3= length of 1st piece in cm  x-4= length of 3rd piece in cm  then since you know they all came from the 20 cm ribbon you add them together and set them equal to 20x+ x-3 +x-4 = 20  3x - 7= 20  3x= 27  x= 9  the shortest piece is the third piece which is x- 4  x-4  9-4 = 5  the length of the smallest piece is 5 cm.  the other lengths are  9 cm  6 cm

What is the diagram u didn't post that