1Cademy - Solving $$2x^2 + 9x - 5 = 0$$ Using the Quadratic Formula

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Quadratic Equation

Example

Solving $2x^2 + 9x - 5 = 0$ Using the Quadratic Formula

Solve $2x^2 + 9x - 5 = 0$ by applying the Quadratic Formula. This example demonstrates the full procedure when the equation is already in standard form and the expression under the square root simplifies to a perfect square.

Step 1 — Identify $a$ , $b$ , $c$ . The equation is already in standard form $ax^2 + bx + c = 0$ . Here $a = 2$ , $b = 9$ , and $c = -5$ .

Step 2 — Substitute into the Quadratic Formula:

$x = \frac{-9 \pm \sqrt{9^2 - 4(2)(-5)}}{2(2)}$

Step 3 — Simplify. Compute inside the square root: $9^2 = 81$ and $4(2)(-5) = -40$ , so $81 - (-40) = 121$ . Since $\sqrt{121} = 11$ :

$x = \frac{-9 \pm 11}{4}$

Split into two solutions:

$x = \frac{-9 + 11}{4} = \frac{2}{4} = \frac{1}{2} \qquad \text{or} \qquad x = \frac{-9 - 11}{4} = \frac{-20}{4} = -5$

Step 4 — Check both solutions:

For $x = \frac{1}{2}$ : $2\left(\frac{1}{2}\right)^2 + 9\left(\frac{1}{2}\right) - 5 = \frac{1}{2} + \frac{9}{2} - 5 = 5 - 5 = 0$ ✓

For $x = -5$ : $2(-5)^2 + 9(-5) - 5 = 50 - 45 - 5 = 0$ ✓

The solutions are $x = \frac{1}{2}$ and $x = -5$ . Because the value under the square root ( $121$ ) turned out to be a perfect square, both solutions are rational numbers. When $c$ is negative, the subtraction $b^2 - 4ac$ involves subtracting a negative product, which increases the value under the radical.

Updated 2026-04-21

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OpenStax Elementary Algebra

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