How do you solve #x^2= 8- 5x #?

1 Answer
Jun 14, 2017

#x = (-5 +- sqrt(57))/2#

Explanation:

The first thing to do is to convert the equation to standard form #y = ax^2 + bx + c#. In this case of our equation, it is #x^2 + 5x - 8 = 0#.

By using this, we know the following
#a = 1#
#b = 5#
#c = -8#

In order to find the value of #x# in the equation above, we need to substitute these values to the quadratic formula #x = (-b +- sqrt(b^2 - 4ac))/ (2a)#

In this case

#x = (-5 +- sqrt(5^2 - 4(1)(-8)))/ (2(1))#
#x = (-5 +- sqrt(25 - (-32)))/2#
#x = (-5 +- sqrt(57))/2#

So we know that #x = (-5 +- sqrt(57))/2#