How do you solve #(2x+3)(x+4)=1 # for x using the square root property? Algebra Properties of Real Numbers Square Roots and Irrational Numbers 1 Answer Manikandan S. Jul 3, 2017 #(2x+3)(x+4)=1# #2x^2+7x+12 =1# #2(x^2+7/2x+49/16)+12-49/8=1# #2(x+7/4)^2+47/8=1# #2(x+7/4)^2=-39/8# #(x+7/4) = \pm sqrt(39/16)i# #x = (-7+- sqrt(39)i)/4# Answer link Related questions What are square roots? How do you simplify square roots that are irrational? How do you find a square root of a decimal? How do you find a square root of a fraction? How do you estimate square roots? What is the square root of 100? What is the square root of 25? What is the square root of 1? How do I know if a square root is irrational? Why are square roots irrational? See all questions in Square Roots and Irrational Numbers Impact of this question 1650 views around the world You can reuse this answer Creative Commons License