Question

How many real solutions does the equation x^2-2x+5=0 have?

Answer 1

Two solutions  x =(2-√-16)/2=1-2i= 1.0000-2.0000i x =(2+√-16)/2=1+2i= 1.0000+2.0000i

Answer 2

There are no real solutions the equation $x^{2} -2x+5=0$ have.

There are two complex solutions the equation $x^{2} -2x+5=0$ have.

How to find real solutions?

If the value of the discriminant($b^{2}-4ac$) is positive, there are two real solutions for x, meaning the graph of the solution has two distinct x-intercepts.

If the value of the discriminant is zero, there is one real solution for x, meaning the graph of the solution has one x-intercept.

If a negative number is inside the square root, there are no x-intercepts.  The solutions are not real numbers, but would have to be expressed as complex numbers.

Given quadratic equation

$x^{2} -2x+5=0$

⇒ $x$ = $\frac{-b}{2a}$ ± $\frac{\sqrt{b^{2} -4ac} }{2a}$

⇒ $x$ = $\frac{2}{2(1)}$ ± $\frac{\sqrt{4-20} }{2(1)}$

⇒ $x$ = 1 ± 2i

Hence, there are two real solutions does the equation $x^{2} -2x+5=0$ have.

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