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

b² + 15b + 56 = 0
(b + 7)(b + 8) = 0

For this expression to be true, the left side of the expression must equal zero. Therefore, either (b + 7) or (b + 8) must equal zero:

If (b + 7) = 0, b must equal -7
If (b + 8) = 0, b must equal -8

So the solution is that b = -7 or -8

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:

x² - 3x - 18 = -x - 3
x² - 3x - 18 + 3 = -x
x² - 3x + x - 15 = 0
x² - 2x - 15 = 0

Next, factor the quadratic equation:

x² - 2x - 15 = 0
(x + 3)(x - 5) = 0

For this expression to be true, the left side of the expression must equal zero. Therefore, either (x + 3) or (x - 5) must equal zero:

If (x + 3) = 0, x must equal -3
If (x - 5) = 0, x must equal 5

So the solution is that x = -3 or 5

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.)

2a(a - x)
2(6)(6 - 5)
2(6)(1)
(12)(1)
12

To solve this equation, isolate the variable for which you are solving (a) on one side of the equation and put everything else on the other side.

-a - 2y = -8a - 6y - 4
-a = -8a - 6y - 4 + 2y
-a + 8a = -6y - 4 + 2y
7a = -4y - 4
a = \( \frac{-4y - 4}{7} \)
a = \( \frac{-4y}{7} \) + \( \frac{-4}{7} \)
a = -\(\frac{4}{7}\)y - \(\frac{4}{7}\)

You need to find the value of c so solve the first equation in terms of x:

5c + x = -5
x = -5 - 5c

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

4c - 3(-5 - 5c) = -1
4c + (-3 x -5) + (-3 x -5c) = -1
4c + 15 + 15c = -1
4c + 15c = -1 - 15
19c = -16
c = \( \frac{-16}{19} \)
c = -\(\frac{16}{19}\)