# Quadratic Equation A8
Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/exercises/quadratic-equation-a-8/
Fetched from algebrica.org test 3482; source modified 2025-03-06T16:12:23.
Solve the [quadratic equation](../quadratic-equations.md):
\\[
x^2 + 0.4x - 0.16 = 0
\\]
* * *
To solve the equation, we can get rid of the decimals by multiplying every coefficient by 100, obtaining:
\\[
100x^2 + 40x - 16 = 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{-40 \\pm \\sqrt{40^2-4(100)(-16)}}{2(100)}\\[0.8em] &= \\frac{-40 \\pm \\sqrt{1600+6400}}{200} \\end{align\*}
* * *
In this case, the discriminant \\(\\Delta\\) is \\(\\geq 0\\) so the equation admits two distinct real solutions. by performing the calculations we have:
\\begin{align\*} x\_{1,2} &= \\frac{{-40\\pm \\sqrt{{8000}}}}{200}\\[0.8em] &= \\frac{{-40\\pm \\sqrt{{1600 \\times 5}}}}{200}\\[0.8em] &= \\frac{{-40\\pm \\sqrt{{40^2 \\times 5}}}}{200}\\[0.8em] &= \\frac{{-40\\pm 40\\sqrt{{5}}}}{200}\\[0.8em] &= \\frac{{-1 \\pm \\sqrt{{5}}}}{5} \\end{align\*}
* * *
Finally, we obtain:
\\[
x\_1 = \\frac{{-1 + \\sqrt{{5}}}}{5}
\\]
\\[
x\_2= \\frac{{-1-\\sqrt{{5}}}}{5}
\\]
The solution to the equation is:
\\[
x = \\frac{{-1+\\sqrt{{5}}}}{5} \\quad x = \\frac{{-1-\\sqrt{{5}}}}{5}
\\]
* * *
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}
\\]