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:

z² + 2z - 14 = -5z + 4
z² + 2z - 14 - 4 = -5z
z² + 2z + 5z - 18 = 0
z² + 7z - 18 = 0

Next, factor the quadratic equation:

z² + 7z - 18 = 0
(z - 2)(z + 9) = 0

For this expression to be true, the left side of the expression must equal zero. Therefore, either (z - 2) or (z + 9) must equal zero:

If (z - 2) = 0, z must equal 2
If (z + 9) = 0, z must equal -9

So the solution is that z = 2 or -9

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

7a + y = 3
y = 3 - 7a

then substitute the result (3 - 7a) into the second equation:

4a + 5(3 - 7a) = -2
4a + (5 x 3) + (5 x -7a) = -2
4a + 15 - 35a = -2
4a - 35a = -2 - 15
-31a = -17
a = \( \frac{-17}{-31} \)
a = \(\frac{17}{31}\)

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.)

-3a(a - x)
-3(-3)(-3 + 8)
-3(-3)(5)
(9)(5)
45

To solve this equation, repeatedly do the same thing to both sides of the equation until the variable is isolated on one side of the equal sign and the answer on the other.

3a + 6 = \( \frac{a}{1} \)
1 x (3a + 6) = a
(1 x 3a) + (1 x 6) = a
3a + 6 = a
3a + 6 - a = 0
3a - a = -6
2a = -6
a = \( \frac{-6}{2} \)
a = -3

To solve this equation, repeatedly do the same thing to both sides of the equation until the variable is isolated on one side of the < sign and the answer on the other.

3x - 2 < \( \frac{x}{-3} \)
-3 x (3x - 2) < x
(-3 x 3x) + (-3 x -2) < x
-9x + 6 < x
-9x + 6 - x < 0
-9x - x < -6
-10x < -6
x < \( \frac{-6}{-10} \)
x < \(\frac{3}{5}\)