how do we solve it

This answer comes from Littlewood's A University Algebra

ax

^{2}+bx+c=0 Multiply by 4a to get4a

^{2}x^{2}+4abx+4ac=0 Compete the square as(2ax+b)

^{2}=4a^{2}x^{2}+4abx+b^{2}so that4a

^{2}x^{2}+4abx+4ac=(2ax+b)^{2}-b^{2}+4ac=0(2ax+b)

^{2}=b^{2}-4ac2ax+b=±√(b

^{2}-4ac)2ax=-b±√(b

^{2}-4ac)x=(-b±√(b

^{2}-4ac))/(2a)