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Final Answer

$$x = \frac{3}{2}\quad\text{or}\quad x = 1$$

How we got there – step‑by‑step using the quadratic formula

The quadratic equation is

$$2x^{2} - 5x + 3 = 0$$

The quadratic formula solves any equation of the form $ax^{2}+bx+c=0$:

$$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$

Identify the coefficients:

• $a = 2$
• $b = -5$
• $c = 3$

Plug them into the formula:

1. Compute the discriminant $\Delta = b^{2}-4ac$

   $$\Delta = (-5)^{2} - 4(2)(3) = 25 - 24 = 1$$

2. Take the square root of the discriminant

   $$\sqrt{\Delta} = \sqrt{1} = 1$$

3. Apply the “±” part

   $$x = \frac{-(-5) \pm 1}{2(2)} = \frac{5 \pm 1}{4}$$

4. Separate the two solutions

   • With the “+” sign: $$x = \frac{5 + 1}{4} = \frac{6}{4} = \frac{3}{2}$$
   • With the “–” sign: $$x = \frac{5 - 1}{4} = \frac{4}{4} = 1$$

So the equation has two real roots: $x = \dfrac{3}{2}$ and $x = 1$.

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Why the quadratic formula works

The formula comes from completing the square on the general quadratic $ax^{2}+bx+c$. It guarantees the exact solutions, even when the roots are irrational or complex. The discriminant $\Delta = b^{2}-4ac$ tells you the nature of the roots:

• $\Delta > 0$: two distinct real roots (as we have here).
• $\Delta = 0$: one repeated real root.
• $\Delta < 0$: two complex conjugate roots.

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A concrete example you can try

Plot the parabola $y = 2x^{2} - 5x + 3$. The points where the curve crosses the $x$-axis are exactly the solutions we found: $(1,0)$ and $\left(\frac{3}{2},0\right)$. Visualizing the graph often makes the connection between the algebraic solution and the shape of the function clearer.

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Want a visual walkthrough?
I’ll suggest a short video that animates the derivation of the quadratic formula and shows how to apply it step by step.

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