# Quadratic Equation A9

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https://algebrica.org/exercises/quadratic-equation-a9/
Fetched from algebrica.org test 3484; source modified 2025-03-06T16:11:49.

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

\\[
7x^2+x+5=0
\\]

* * *

The equation is already reduced to the standard form \\(ax^2+bx+c= 0\\). We can substitute the coefficients \\(a=2, b=10, c=11\\) into the [quadratic formula](../quadratic-formula.md):

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

We obtain:

\\begin{align\*} x\_{1,2} &= \\frac{{-(1) \\pm \\sqrt{{(1)^2 - 4(7)(5)}}}}{{2(7)}}\\[0.8em] &= \\frac{{-1 \\pm \\sqrt{{1 - 140)}}}}{{14}}\\[0.8em] &= \\frac{{-1 \\pm \\sqrt{{-139}}}}{{14}}\\\\ \\end{align\*}

* * *

In this case, the discriminant \\(\\Delta\\) is \\(\\leq 0\\) which means the equation has no real solutions.

The solution to the equation is:

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

* * *

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}
\\]