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(-4)(-4 - 8)
7(-4)(-12)
(-28)(-12)
336

The first step to solve a quadratic equation that's set to zero is to factor the quadratic equation:

x² + 3x - 18 = 0
(x - 3)(x + 6) = 0

For this expression to be true, the left side of the expression must equal zero. Therefore, either (x - 3) or (x + 6) must equal zero:

If (x - 3) = 0, x must equal 3
If (x + 6) = 0, x must equal -6

So the solution is that x = 3 or -6

To combine like terms, add or subtract the coefficients (the numbers that come before the variables) of terms that have the same variable raised to the same exponent.

7a - 2a = 5a

You need to find the value of a so solve the first equation in terms of z:

+ z = 5
z = 5 +

then substitute the result (5 - 0a) into the second equation:

9a - 9(5 + ) = -6
9a + (-9 x 5) + (-9 x ) = -6
9a - 45 + = -6
9a + = -6 + 45
9a = 39
a = \( \frac{39}{9} \)
a = 4\(\frac{1}{3}\)

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 = \( \sqrt{a} \) = \( \sqrt{9} \) = 3

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 = 3² + 3²
c² = 18
c = \( \sqrt{18} \) = \( \sqrt{9 x 2} \) = \( \sqrt{9} \) \( \sqrt{2} \)
c = 3\( \sqrt{2} \)