How do you find the solution to the quadratic equation #2x^2 + 7 = 3x#? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer sankarankalyanam Jun 18, 2018 Answer: #color(maroon)(x_+ = (3 + i sqrt(47)) / 4, x_- = (3 - i sqrt(47))/4# Explanation: #2x^2 + 7 = 3x# #2x^2 - 3x + 7 = 0# #a = 2, b = -3, c = 7# #x = (-(-3) +- sqrt((-3)^2 - (4 * 2 * 7)))/(2 * 2)# #x = (3 +- sqrt(9 - 56) ) / 4# #x = (3 +- sqrt(-47)) / 4# #color(maroon)(x_+ = (3 + i sqrt(47)) / 4, x_- = (3 - i sqrt(47))/4# 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 504 views around the world You can reuse this answer Creative Commons License