How do you solve $x^2 – 5x – 1 = - 7$ using the quadratic formula?

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Answer 1 · 2016-05-13T05:47:21

$x=(5+-sqrt(-23))/4$

It has imaginary roots

Given -

$x^2-5x-1=-7$

Take the constant to the LHS.

$2x^2-5x-1+7=0$
$2x^2-5x+6=0$

$x=(-b+-sqrt(b^2-4ac))/(2a)$
$x=(-(-5) +- sqrt((-5)^2-(4xx2xx6)))/(2xx2)$
$x=(5+-sqrt(25-48))/4$
$x=(5+-sqrt(-23))/4$

It has imaginary roots

How do you solve x^2 – 5x – 1 = - 7x^2 – 5x – 1 = - 7 using the quadratic formula?