x² +x += 0

x = 2 or x = 3

As decimals: x = 2 · x = 3

Equationx² - 5x + 6 = 0
Discriminant b² − 4ac1 (two real roots)
Vertex(2.5, -0.25)
Axis of symmetryx = 2.5
y-intercept(0, 6)
Show the working
  1. Start from the quadratic formula: x = (−b ± √(b² − 4ac)) / 2a
  2. Substitute a = 1, b = -5, c = 6: x = (5 ± √(-5² − 4·1·6)) / 2
  3. Work out the discriminant: -5² − 4·1·6 = 1
  4. Take the square root and divide by 2 to get the roots above.

Exact answers, not just decimals

Most online solvers hand back a rounded decimal. When a quadratic does not factor neatly the honest answer is a surd — for x² − 3x + 1 = 0 the roots are (3+√5)/2 and (3−√5)/2, and writing 2.618 instead will cost marks in most exam boards. This page runs the same symbolic engine as our TI-89 workspace, so it reduces the surd properly and only falls back to decimals when there is nothing exact to give.

How the calculator does it

The TI-84 has no dedicated quadratic program, so there are two standard routes and both work in our simulator:

Two ways to solve a quadratic on a TI-84
MethodKeystrokesWhen to use it
Numerical solverMATH → B: Solver, giving solve(X²-5X+6,X,1)Fast when you only need one root and can guess near it
Graph and find the zerosY=, GRAPH, then 2nd + TRACE → 2: zeroBest when you need both roots, or want to see the shape

The full keystroke reference for both menus is in the TI-84 manual, and the graphing guide walks through the bound prompts that 2: zero asks for.

Frequently asked questions

What does the discriminant tell me?

The discriminant is b² − 4ac, the part under the square root. If it is positive there are two real roots, if it is zero there is one repeated root, and if it is negative the parabola never meets the x-axis so there are no real roots.

Why are the answers shown as surds instead of decimals?

Because that is usually what the question asks for. A root such as (3+√5)/2 is exact, while 2.618033988 is rounded. Both are shown here, so you can quote whichever your teacher expects.

How do I solve a quadratic on the TI-84 itself?

Two ways. Type solve(X²-5X+6,X,1) on the home screen for a numerical root near your guess, or enter the expression in Y=, press GRAPH, then use 2nd TRACE → 2:zero and answer the Left Bound, Right Bound and Guess prompts.

What if a is zero?

Then the x² term disappears and the equation is linear, not quadratic. This solver detects that and solves bx + c = 0 instead, which has a single root at x = −c/b.

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