How do you solve #20x^2 = 13 - x#?

1 Answer

Answer:

#x~~0.78,0.83#

Explanation:

Rearranging the equation
#=>20x^2+x-13=0 => (1)#
#ax^2+bx+c=0##=>#standard equation

Comparing (#1#) with standard equation

#a=20,b=1,c=-13#

Using quadratic formula:

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

#=>(-1+-sqrt(1^2-4(20×-13)))/(2(20))#

#=>(-1+-sqrt(1-4(-260)))/40#

#=>(-1+-sqrt(1+1040))/40#

#=>(-1+-sqrt(1041))/40#

#=>(-1+-32.26)/40#

#x~~(-1+32.26)/40,(-1-32.26)/40#

#x~~31.26/40,-33.26/40#

#x~~0.78,0.83#