How do you solve this quadratic equation #2x^2- 10+ 7=0#?

2 Answers
Jun 3, 2018

Answer:

#x=+-1/2sqrt6#

Explanation:

#2x^2-10+7=0#
Add 3 on both sides to move the constants to the right side:
#2x^2-10+7+3=0+3#
#2x^2=3#
Divide both sides with 2:
#x^2=3/2#
#x=+-sqrt(3/2)=+-1/2sqrt3sqrt2=+-1/2sqrt6#
(if we multiply the right side with #sqrt2/sqrt2#)

Jun 3, 2018

Answer:

# x = (5 pm sqrt(11))/2 #

Explanation:

if #2x^2 -10x + 7 = 0 #

You can use the quadratic formula:

#ax^2 +bx + c = 0 #

#=> x = ( -bpmsqrt(b^2 - 4ac) ) / (2a) #

In this case #a = 2 # , #b = -10 # , # c = 7 #

#=> x =( - (-10 ) pm sqrt( (-10)^2 - (4 * 2 * 7 ) ) ) / ( 2 * 2) #

#=> x = (10 pm sqrt(44)) / 4 #

#=> x= (10 pm 2sqrt(11) )/ 4 #

#color(red)(=> x = (5 pm sqrt(11))/2 #