# Quadratic Equation A4

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https://algebrica.org/exercises/quadratic-equation-a-4/
Fetched from algebrica.org test 3464; source modified 2025-03-06T16:13:44.

Solve the [quadratic equation](../quadratic-equations.md):

\\[
x^2-5x-14 = 0
\\]

* * *

The equation is already reduced to the standard form \\(ax^2+bx+c= 0\\). We can substitute the coefficients \\(a=1, b=-5, c=-14\\) into the [quadratic formula](../quadratic-formula.md):

\\[
x\_{1,2} = \\frac{{-b \\pm \\sqrt{{b^2-4ac}}}}{{2a}}
\\]

We obtain:

\\[
x\_{1,2} = \\frac{{-(-5) \\pm \\sqrt{{(-5)^2-4(1)(-14)}}}}{{2(1)}}
\\]

* * *

In this case, the discriminant \\(\\Delta\\) is \\(\\geq 0\\) so the equation admits two distinct real solutions.

\\begin{align\*} x\_{1,2} &= \\frac{{5 \\pm \\sqrt{{25+56}}}}{2}\\[0.6em] &= \\frac{{5 \\pm \\sqrt{{81}}}}{2}\\\\ \\end{align\*}

* * *

Finally, by performing the calculations, we obtain:

\\[
x\_1 = \\frac{5+ 9}{2} = \\frac{14}{2} = 7
\\]



\\[
x\_2=\\frac{5-9}{2} = -\\frac{4}{2} = -2
\\]

The solution to the equation is:

\\[
x =7 \\quad \\quad x=-2
\\]

* * *

Remember that the [discriminant](../quadratic-formula.md) is crucial in determining the nature and number of solutions of quadratic equations.

-   If \\( b^2 - 4ac > 0\\), the quadratic equation has two distinct real solutions.

\\[
S = \\{x\_1, x\_2\\} \\quad x\_1, x\_2 \\in \\mathbb{R} \\quad x\_1 \\neq x\_2
\\]


-   If \\( b^2 - 4ac = 0\\), the quadratic equation has two coincident real solutions.

\\[
S = \\{x\\} \\quad x \\in \\mathbb{R} \\quad x = x\_1 = x\_2
\\]


-   If \\( b^2 - 4ac = < 0\\), the quadratic equation has no real solutions. Instead, it gives rise to complex solutions characterized by imaginary components.

\\[
\\nexists \\hspace{10px} x \\in \\mathbb{R}
\\]