(x+1)(4x+1)=23x
(4x+1)2=3x(x+1)
8x+2=3x2+3x
3x2−5x−2=0
Using the quadratic formula x=2a−b±b2−4ac with a=3, b=−5, and c=−2:
x=2(3)−(−5)±(−5)2−4(3)(−2)
x=65±25+24
x=65±49
x=65±7
This gives two possible solutions for x:
The possible values of x are 2 and −31.