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(D) The given equation is $x^{2}+2 \sqrt{2} x-6=0$.Comparing this with the standard form $ax^{2}+bx+c=0$,we get $a=1, b=2 \sqrt{2}$,and $c=-6$.The qua

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$\sqrt{2},-3 \sqrt{2}$

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(D) The given equation is $x^{2}+2 \sqrt{2} x-6=0$.Comparing this with the standard form $ax^{2}+bx+c=0$,we get $a=1, b=2 \sqrt{2}$,and $c=-6$.The quadratic formula is $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$.Substituting the values,we get $x = \frac{-(2 \sqrt{2}) \pm \sqrt{(2 \sqrt{2})^{2}-4(1)(-6)}}{2(1)}$.$x = \frac{-2 \sqrt{2} \pm \sqrt{8+24}}{2} = \frac{-2 \sqrt{2} \pm \sqrt{32}}{2}$.Since $\sqrt{32} = 4 \sqrt{2}$,we have $x = \frac{-2 \sqrt{2} \pm 4 \sqrt{2}}{2}$.Calculating the two roots:$x_1 = \frac{-2 \sqrt{2} + 4 \sqrt{2}}{2} = \frac{2 \sqrt{2}}{2} = \sqrt{2}$.$x_2 = \frac{-2 \sqrt{2} - 4 \sqrt{2}}{2} = \frac{-6 \sqrt{2}}{2} = -3 \sqrt{2}$.Thus,the roots are $\sqrt{2}$ and $-3 \sqrt{2}$.

Find the roots of the quadratic equation using the quadratic formula:
$x^{2}+2 \sqrt{2} x-6=0$

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Find the roots of the quadratic equation using the quadratic formula:$x^{2}+2 \sqrt{2} x-6=0$