Answer:
Impossible
Step-by-step explanation:
In 5x^2-4x+3=0, standard form, substitute these values in the quadratic formula:
a = 5; b = -4; c = 3
The quadratic formula is [tex]x = \frac{ -b ± \sqrt{b^{2} - 4ac}}{2a}[/tex]
(ignore the weird capital A)
Substitute a b and c:
[tex]x = \frac{-(-4) ± \sqrt{(-4)^{2} - 4(5)(3)}}{2(5)}[/tex]
Simplify:
[tex]x = \frac{4) ± \sqrt{16 - 60}}{10}[/tex]
Because [tex]\sqrt{16-60}[/tex] is the square root of a negative number, the answer would be imaginary.
Therefore, there are not solutions to this equation.
A solution is the same as the roots or zeroes, where the graph would cross the x-axis when graphed.
The graph never meets the x-axis. It looks like this:
