Question

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Answer 1

Answer:

Option A

Step-by-step explanation:

We know that for an eqn. $ax^{2} +bx+c=0$ , the quadratic formula is =

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

So using quadratic eqn. for $x^{2} -6x-5=0$

$x =\frac{-(-6)+\sqrt{(-6)^{2}-4*1*(-5) } }{2*1}or\frac{-(-6)-\sqrt{(-6)^{2}-4*1*(-5) } }{2*1}$

$=> x = \frac{6+\sqrt{36+20} }{2} or\frac{6-\sqrt{36+20} }{2}$

$=>x = \frac{6+\sqrt{56} }{2} or\frac{6-\sqrt{56} }{2}$