How do you find the solution to the quadratic equation # (x - 5)^2 + 2(x - 5) - 35 = 0 #?

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Answer 1 · 2015-05-01T19:20:47

Easier if you let #p=x-5#,
solve the equation for "p" ,
then substitute back to get the values for #x#

#(x-5)^2+2(x-5)-35=0#

#p^2+2p-35=0#

#(p+7)(p-5) = 0#

If #p+7=0#
then
#x-5+7 = 0#
#x=-2#

If #p-5=0#
then
#x-5-5=0#
#x=10#

The solutions for the given quadratic equation are
#x=-2 " and " x=10#