Question

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

Answer 1

$\bf x^2=4x-5\implies x^2-4x+5=0\\\\ -------------------------------\\\\ \begin{array}{llccll} & 1x^2& -4x& +5&=&0\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array}\qquad \qquad (-4)^2-4(1)(5) \\\\\\ discriminant\implies b^2-4ac= \begin{cases} 0&\textit{one solution}\\ positive&\textit{two solutions}\\ negative&\textit{no solution} \end{cases}$