SOLUTION: tell if the equation has two solutions, one solution, or no real solution: {{{x^2-3x+4=0}}}

Question 810574: tell if the equation has two solutions, one solution, or no real solution:
Answer by Charles3475(23) (Show Source): You can put this solution on YOUR website!
In algebra, the discriminant of a polynomial is a function of its coefficients, typically denoted Greek letter Delta (Δ).The discriminant of the quadratic polynomial
ax^2+bx+c
is
Δ=b^2-4ac
A quadratic polynomial with real coefficients has real roots (solutions when the function is equal to zero) if and only if the discriminant is non-negative, and these roots are distinct if and only if it is positive (not zero). Thus
Δ > 0: 2 distinct real roots
Δ < 0: 2 distinct complex roots
Δ = 0: 1 real root with multiplicity 2
A complex root is not a real root. The square root of negative 1 (i=√(-1)) is used for all complex numbers. This number is called "i", standing for "imaginary", because i is not "real".
For your question: tell if the equation has two solutions, one solution, or no real solution:
a=1
b=-3
c=4
The discriminant equals b^2-(4)(a)(c) = (-3)^2-(4)(1)(4) = 9 - 16 = -7
As the discriminant is negative there are no real roots.
There are two complex roots.
All of this falls out of the quadratic formula. The discriminant is part of the formula. Take a look!
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