How do you solve #x + 9 = 2(x - 1)^2#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Black Widow Apr 15, 2018 #x_1=7/2# #x_2=-1# Explanation: #x+9=2(x^2-2x+1)# #x+9=2x^2-4x+2# #-2x^2+x+4x+9-2=0# #-2x^2+5x+7=0 //*(-1)# #2x^2-5x-7=0# Now use the formula #x_(1//2)=(-b+-sqrt(b^2-4ac))/(2a)# #a=2# #b=-5# #c=-7# #x_(1//2)=(5+-sqrt((25-4*2(-7))))/(2*2)# #x_(1//2)=(5+-sqrt((25+56)))/4# #x_(1//2)=(5+-sqrt(81))/4# #x_(1//2)=(5+-9)/4# #x_1=(5+9)/4=14/4=7/2# #x_2=(5-9)/4=(-4)/4=-1# Answer link Related questions What are the different methods for solving quadratic equations? What would be the best method to solve #-3x^2+12x+1=0#? How do you solve #-4x^2+4x=9#? What are the two numbers if the product of two consecutive integers is 72? Which method do you use to solve the quadratic equation #81x^2+1=0#? How do you solve #-4x^2+4000x=0#? How do you solve for x in #x^2-6x+4=0#? How do you solve #x^2-6x-16=0# by factoring? How do you solve by factoring and using the principle of zero products #x^2 + 7x + 6 = 0#? How do you solve #x^2=2x#? See all questions in Comparing Methods for Solving Quadratics Impact of this question 1438 views around the world You can reuse this answer Creative Commons License