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What are the solutions of the quadratic equation 2x^(2)-5x+3=0.

What are the solutions of the quadratic equation $2x^{2}-5x+3=0$ .

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Q. What are the solutions of the quadratic equation

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

.

Identify the quadratic equation: Identify the quadratic equation. $\newline$ The given equation is $2x^2 - 5x + 3 = 0$ , which is a standard form of a quadratic equation $ax^2 + bx + c = 0$ , where $a = 2$ , $b = -5$ , and $c = 3$ .
Apply the quadratic formula: Apply the quadratic formula to find the roots of the equation. $\newline$ The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . We will use this formula to find the roots of the equation $2x^2 - 5x + 3 = 0$ .
Calculate the discriminant: Calculate the discriminant $b^2 - 4ac$ . $\newline$ The discriminant is the part of the quadratic formula under the square root, $b^2 - 4ac$ . For our equation, $b = -5$ , $a = 2$ , and $c = 3$ . $\newline$ Discriminant = $(-5)^2 - 4(2)(3) = 25 - 24 = 1$ .
Calculate the roots: Calculate the roots using the quadratic formula. $\newline$ Now that we have the discriminant, we can find the two roots of the equation. $\newline$ Root $1$ : $x = \frac{-(-5) + \sqrt{1}}{2 \times 2} = \frac{5 + 1}{4} = \frac{6}{4} = 1.5$ $\newline$ Root $2$ : $x = \frac{-(-5) - \sqrt{1}}{2 \times 2} = \frac{5 - 1}{4} = \frac{4}{4} = 1$

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