How do you solve #x^{2}-6x=26#?

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Answer 1 · 2017-10-21T15:47:46

# x = (3 + sqrt(35)), (3 - sqrt(35))#

# x = 8.9161, - 2.9161#

#x^2 -6x -26 = 0#

Equation is in the form #ax^2 + bx + c = 0#.
The roots are # (-b +- (sqrt (b^2 -4ac))/(2a))#

#a = 1, b = -6, c = -26#

#:. x = 6 +- sqrt (6^2 -( 4* 1* (-26))) / 2#

#x = (6 +- sqrt( 36 + 104))/2 = (6 +- sqrt140)/2#

#x = cancel(2)( (3 +- sqrt35) / cancel(2))#

# x = (3 + sqrt(35)), (3 - sqrt(35))#

# x = 8.9161, - 2.9161#