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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