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Solve using quadratic formula
5x^2-4x+3=0

User Riwall
by
4.7k points

1 Answer

3 votes

Answer:

Impossible

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
x = \frac{ -b ± \sqrt{b^(2) - 4ac}}{2a}

(ignore the weird capital A)

Substitute a b and c:


x = \frac{-(-4) ± \sqrt{(-4)^(2) - 4(5)(3)}}{2(5)}

Simplify:


x = (4) ± √(16 - 60))/(10)

Because
√(16-60) 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:

Solve using quadratic formula 5x^2-4x+3=0-example-1
User Benske
by
5.2k points