Solve with the quadratic formula:

the length of a rectangle is 1 less than 2 times the width. the area of the rectangle is 45 cm^2. find the length and width.

Let the width be x.
Length = 2x -1

The area of rectangle = 45 cm²

x*(2x -1) = 45

2x² -x = 45

2x² -x - 45 = 0              This is a quadratic equation

comparing to ax² + bx + c = 0,      a = 2, b = -1, c = -45

x =  (-b + √(b² -4ac)) / 2a      or   (-b - √(b² -4ac)) / 2a

x =  (- -1 + √((-1)² -4*2*-45)) / 2*2      or   (- -1 - √((-1)² -4*2*-45)) / 2*2

x =   (1 + √(1 +360)) / 2*2      or   (1 - √(1 + 360) / 2*2

x =   (1 + √361) / 4     or   (1 - √361) / 4

x =   (1 + 19) / 4     or   (1 - 19) / 4

x =  20/4  or  -18/4

x = 5 or -4.5

x can't be negative since we are solving for side.

x = 5 as the only valid solution.

Recall, the width = x = 5.
Length = 2x - 1 = 2*5 - 1 = 10 - 1 = 9

Hence the length = 9cm , and width = 5cm

If is congruent to,then m is the midpoint