How do you solve #2x^2 = 14x + 20#?

2 Answers
Apr 20, 2018

#x = (7 +- sqrt(89))/2#

Explanation:

#=> 2x^2 = 14x + 20#

Divide both sides by #2#

#=> x^2 = 7x + 10#

Rearrange the equation

#=> x^2 - 7x - 10 = 0#

It’s in the form of #ax^2 + bx + c = 0#

where,

  • #a = 1#
  • #b = -7#
  • #c = -10#

Use formula for quadratic equation to find #x#

#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

#x = (-(-7) +- sqrt((-7)^2 - (4 × 1 × -10)))/(2 × 1)#

#x = (7 +- sqrt(49 + 40))/2#

#x = (7 +- sqrt(89))/2#

Apr 20, 2018

#(7\pm\sqrt(89))/2#

Explanation:

  • Move everything to left side: #2x^2-14x-20=0#
  • Factor out a 2: #2(x^2-7x-10)=0#
  • Solve the parenthetical quadratic (see here for steps)