Question

find the zeros of the quadratic equation :-
2x^2-12x + 4=0​

Answer 1

Answer:

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Step-by-step explanation:

Hence, the roots of the given equation are 3 & ½.

Answer 2

$2 {x}^{2} - 12x + 40 = 0 \\ = > 2( {x}^{2} - 6x + 20) = 0 \\ = > {x}^{2} - 6x + 20= 0 \\ = > x = \frac{ - ( - 6) + \sqrt{ {( - 6)}^{2} - 4 \times 20 \times 1}}{2 \times 1} \\ = \frac{6 + \sqrt{ - 44} }{2} \\ = \frac{6 + 2 \sqrt{ - 11} }{2 } \\ = 3 + \sqrt{ - 11} \\ \\ \\ \\ \\ \\ or \: \: \: \: x = \frac{ - ( - 6) - \sqrt{ {( - 6)}^{2} - 4 \times 20 \times 1}}{2 \times 1} \\ = \frac{6 - \sqrt{ - 44} }{2} \\ = \frac{6 - 2 \sqrt{ - 11} }{2 } \\ = 3 - \sqrt{ - 11}$

So, the roots are (3+√-11) and (3-√-11)