How do you solve #x^2 – 3x = 7x – 2# using the quadratic formula?

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Answer 1 · 2015-07-17T04:04:49

#x=5+sqrt(23), x=5-sqrt(23)#

#x^2-3x=7x-2#

Get all of the terms on the left side.

#x^2-3x-7x+2=0# =

#x^2-10x+2=0#

The equation is now in the form of a quadratic equation #ax^2+bx+c#, where #a=1,# #b=-10,# and #c=2#.

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

Substitute the values for #a, b, and c# into the formula.

#x=(-(-10)+-sqrt(-10^2-4*1*2))/(2*1)# =

#x=(10+-sqrt(100-8))/(2)# =

#x=(10+-sqrt(92))/2#

Simplify #sqrt(92)#.

#sqrt(92)=sqrt(2xx2xx23)# =

#sqrt(92)=2sqrt23#

Substitute #2sqrt(23)# for #sqrt(92)#.

#x=(10+-2sqrt(23))/2#

Simplify.

#x=5+-sqrt(23)#

Solve for #x#.

#x=5+sqrt(23)#

#x=5-sqrt(23)#