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

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Answer 1 · 2016-12-11T02:50:05

#x=1+sqrt7,##1-sqrt7#

#x^2-2x=6#

Subtract #6# from both sides.

#x^2-2x-6=0# is a quadratic equation, #ax^2+bx+c#, where #a=1#, #b=-2#, and #c=-6#.

If by evaluate you mean solve, use the quadratic formula.

Quadratic Formula

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Substitute the known values into the formula and solve.

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

#x=(2+-sqrt(4+24))/2#

#x=(2+-sqrt(28))/2#

#x=(2+-sqrt(2xx2xx7))/2#

#x=(2+-2sqrt7)/2#

#x=(cancel(2)^1+-sqrt7)/cancel(2)^1#

#x=1+sqrt7,##1-sqrt7#