The first step to solve a quadratic expression that's not set to zero is to solve the equation so that it is set to zero:

b² + 3b + 6 = -b + 3
b² + 3b + 6 - 3 = -b
b² + 3b + b + 3 = 0
b² + 4b + 3 = 0

Next, factor the quadratic equation:

b² + 4b + 3 = 0
(b + 1)(b + 3) = 0

For this expression to be true, the left side of the expression must equal zero. Therefore, either (b + 1) or (b + 3) must equal zero:

If (b + 1) = 0, b must equal -1
If (b + 3) = 0, b must equal -3

So the solution is that b = -1 or -3

The surface area of a cube is (2 x length x width) + (2 x width x height) + (2 x length x height):

sa = 2lw + 2wh + 2lh
sa = (2 x 1 x 9) + (2 x 9 x 2) + (2 x 1 x 2)
sa = (18) + (36) + (4)
sa = 58

To solve this equation, replace the variables with the values given and then solve the now variable-free equation. (Remember order of operations, PEMDAS, Parentheses, Exponents, Multiplication/Division, Addition/Subtraction.)

7a(a - y)
7(-2)(-2 + 7)
7(-2)(5)
(-14)(5)
-70

A cube is a rectangular solid box with a height (h), length (l), and width (w). The volume is h x l x w and the surface area is 2lw x 2wh + 2lh.

To find the diagonal we need to know the length of one of the square's sides. We know the area and the area of a square is the length of one side squared:

a = s²

so the length of one side of the square is:

s = \( \sqrt{a} \) = \( \sqrt{25} \) = 5

The Pythagorean theorem defines the square of the hypotenuse (diagonal) of a triangle with a right angle as the sum of the squares of the other two sides:

c² = a² + b²
c² = 5² + 5²
c² = 50
c = \( \sqrt{50} \) = \( \sqrt{25 x 2} \) = \( \sqrt{25} \) \( \sqrt{2} \)
c = 5\( \sqrt{2} \)