Hi there i just have about 4 questions i couldn't solve in math

these questions are word problems related to quadratic equations

q1/ the length of a rectangle is one less than twice the width. find the length of the diagonal of the rectangle if the area of the rectangle is 120 units squared.

q2/ one side of the right triangle is 3 more than 3 times the other. the hypotenuse is 3 less than 4 times the shortest side. find the perimeter of the triangle.

q3/ A triangle has a base that is 2 more than twice the height. find the length of the base for the triangle if the area is 30 square units.

q4 -2(4x-1)^2+18=0

1)

Let the width of the rectangle be x units.
The length of the rectangle is 2x - 1 units.

Thus, we have:
x(2x - 1) = 120
2x^2 - x = 120
2x^2 - x - 120 = 0
2x^2 - 16x + 15x - 120 = 0
(2x^2 - 16x) + (15x - 120) = 0
(2x)(x - 8) + 15(x - 8) = 0
(2x + 15)(x - 8) = 0
x = -15/2 or x = 8

Since the width cannot be negative, we reject x = -15/2.
Thus, x = 8.

Length of diagonal
= sqrt((length of rectangle)^2 + (width of rectangle)^2)
= sqrt((2x - 1)^2 + x^2)
= sqrt((2(8) - 1)^2 + 8^2)
= sqrt(15^2 + 64)
= sqrt(225 + 64)
= sqrt(289)
= 17 units

2)

Let the shortest side be x units.
The other side will be 3x + 3 units.
The hypotenuse will be 4x - 3 units.

Perimeter of the triangle
= x + (3x + 3) + (4x - 3)
= 8x units

(Since you did not give me additional information, I will say that the perimeter of the triangle could be anything, since x could be any value.)

3)

Let the height of the triangle be x units.
The base will be 2x + 2 units.

Thus, we have:
Area of triangle = 30
(1/2)(2x + 2)(x) = 30
x^2 + x = 30
x^2 + x - 30 = 0
(x + 6)(x - 5) = 0
x = -6 or x = 5

Since the height cannot possibly be negative, we reject x = -6
Thus, x = 5.

Base of triangle
= 2x + 2 units
= 2(5) + 2 units