Solve for #x# in the equation? #x^2+12x=45#

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Answer 1 · 2018-03-06T18:50:25

#x=-15 or x=3#

There is an #x^2# term, therefore this is a quadratic equation.

#x^2 +12 x-45 =0" "larr#Make it equal to #0#

Factorise: Find factors of #45# which subtract to give #3#

#" " 15 xx 3 = 45 " " and 15-3=12#

#(x+15)(x-3)=0#

Set each factor equal to #0# and solve to find #x#

#x+15 =0" "rarr x = -15#

#x-3 = 0" "rarr x =3#

These are the two solutions,