consider the quadratic equation x^2=4x-5.How many solutions does the equation have?

Hello from MrBillDoesMath! Answer: 2 (complex) solutions    x = 2 + i and x 2 - i
Discussion: Rewrite x^2 = 4x - 5 as x^2 -4x + 5 = 0The solutions from the quadratic formula arex = (  -(-4) +\-  sqrt ( (-4)^2 - 4 (1)(5))  ) /2x = (   4    +\- sqrt(16-20))  )/2x = (  4    +\-  sqrt(-4) ) /2x = (4 +\- 2i) /2x = 2 +\- i

Thank you, MrB