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$7x^2-5x+13=0$

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Multiply and divide rational numbers

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Q.

7x^2-5x+13=0

Calculate Discriminant: To determine the number of real solutions for the quadratic equation $7x^2-5x+13=0$ , we can use the discriminant method. The discriminant ( $D$ ) is given by the formula $D = b^2 - 4ac$ , where $a$ , $b$ , and $c$ are the coefficients of the quadratic equation $ax^2 + bx + c = 0$ .
Find Coefficients: For the given equation $7x^2-5x+13=0$ , the coefficients are: $a = 7$ , $b = -5$ , and $c = 13$ . Let's calculate the discriminant $(D)$ . $\newline$ $D = (-5)^2 - 4(7)(13)$
Perform Calculations: Now, we perform the calculations: $\newline$ $D = 25 - 4(7)(13)$ $\newline$ $D = 25 - 364$
Subtract Numbers: Subtracting $364$ from $25$ gives us: $\newline$ $D = -339$
Interpret Results: Since the discriminant $D$ is negative ( $D = -339$ ), this means that the quadratic equation $7x^2-5x+13=0$ has no real solutions. It has two complex solutions instead.

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