How do you solve $x^2-6x-5=0$ using the quadratic formula?

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Answer 1 · 2016-02-29T10:20:38

x= $3+-sqrt(14)$

Solution of quadratic equation $ax^2 +bx +c=0$ is,

x= $(-b+-sqrt(b^2-4ac))/(2a)$ . This is known as the quadratic formula.

The solution of the given equation would be x= $(6+-sqrt(36+20))/2$

x= $3+-sqrt(14)$

How do you solve x^2-6x-5=0x^2-6x-5=0 using the quadratic formula?