How do I use the quadratic formula without losing track of its signs?

Start by rewriting the equation in standard form, ax² + bx + c = 0. Then label the coefficient of x² as a, the coefficient of x as b, and the constant as c. Keep the signs that belong to the coefficients: for example, if the x-term is −5x, then b = −5, not 5. The formula is x = (−b ± √(b² − 4ac))/(2a). Putting parentheses around each substituted coefficient helps prevent sign errors, particularly with negative values. [S1]

A useful scratch line is:

a = [signed value], b = [signed value], c = [signed value]

In x² − 5x + 6 = 0, for example, the coefficients are a = 1, b = −5, and c = 6.

Take x² − 5x + 6 = 0. With a = 1, b = −5, and c = 6, write the values in parentheses before simplifying:

x = (−(−5) ± √((−5)² − 4(1)(6))) / (2(1))

Now simplify in small steps. The first term, −(−5), is 5. The squared term is 25, and 4(1)(6) is 24. So:

x = (5 ± √(25 − 24))/2 = (5 ± 1)/2

The parentheses make it visible that the formula’s −b is acting on the whole value b = −5. This approach follows the formula’s instruction to substitute a, b, and c, and is especially useful when a coefficient is negative. [S1]

Treat ± as two separate calculations: one using plus and one using minus. For the example, keep the denominator the same in both:

1. Plus case: x = (5 + 1)/2 = 6/2 = 3.
2. Minus case: x = (5 − 1)/2 = 4/2 = 2.

Write both results down; do not stop after calculating the plus case. The ± in the quadratic formula represents these two possible values, so simplifying each case separately keeps the arithmetic organized. [S1]

For your own problem, copy this template:

x = ([−b value] + √[discriminant]) / [2a value] = [first root] x = ([−b value] − √[discriminant]) / [2a value] = [second root]

Here, “discriminant” is a label for the expression b² − 4ac while you work. Keep the coefficient substitutions visible until you have simplified that expression.

Substitute each proposed root into the original equation, keeping the terms and signs as written. For x² − 5x + 6 = 0:

For x = 3: 3² − 5(3) + 6 = 9 − 15 + 6 = 0. For x = 2: 2² − 5(2) + 6 = 4 − 10 + 6 = 0.

Both checks give zero, so both values satisfy this equation. To check your own answers, replace every x in the original equation with one candidate at a time, calculate the left side, and see whether it equals zero. Substituting proposed roots back into the original equation is the recommended check. [S1] A check can reveal an arithmetic or sign mistake, but it does not show which earlier step caused it; retrace the substitution if a result is not zero.