How to Solve Linear and Quadratic Equations

Verify an equation result by substituting it into the original expression.

Use the Equation Solver alongside this illustrative example.

A practical example

For x² − 5x + 6 = 0, enter a = 1, b = −5 and c = 6. The roots are 2 and 3: substituting 2 gives 4 − 10 + 6 = 0, and substituting 3 gives 9 − 15 + 6 = 0. The negative sign on b is essential.

Check the result

Move every term to the same side before entering coefficients. If a is zero, the problem is linear rather than quadratic. A negative discriminant requires complex roots rather than a claim that no solution exists at all.

Try a comparison

Change the constant from 6 to 4 while retaining a = 1 and b = −5. The roots become 1 and 4. Their sum remains 5 while their product changes to 4, giving two independent checks of which coefficient you changed.

What this does not establish

Decimal rounding may leave a small residual on substitution. Keep more digits for that check and distinguish approximate numeric output from exact symbolic work.

Report unclear guidance or a result that needs checking.