# Quadratic Equation B10
Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/exercises/quadratic-equation-b-10/
Fetched from algebrica.org test 5012; source modified 2025-03-06T18:08:06.
Solve the [quadratic equations](../quadratic-equations.md) using the [factorization method](../factoring-quadratic-equations.md).
\\[
12x^2 +17x - 5 = 0
\\]
* * *
The equations is in the standard form \\(ax^2+bx+c=0\\). It is essential to verify the its [discriminant](../quadratic-formula.md) \\(\\Delta = b^2 - 4ac\\) is \\(\\geq0\\) to ensure the equation admits solutions in the field of real numbers. Substituting the coefficients of the equation into \\(\\Delta\\), we get:
\\[
\\Delta = (17)^2 - 4(12)(-5) = 529 \\geq 0
\\]
\\(\\Delta \\gt 0\\) means the equation has real solutions.
* * *
Now, we need to [factorize](../factoring-quadratic-equations.md) the [polynomial](../polynomials.md). We must find two numbers, \\(r\_1, r\_2\\) whose sum \\(S = r\_1 + r\_2\\) equals \\(b = 17\\) and whose product \\(P = r\_1 \\cdot r\_2\\) equals \\(a \\cdot c = 12 \\cdot -5= -60\\). We can use this simple table to find the numbers that satisfy our constraints. \\(-60\\) can be factored into 20 and - 3 and that the sum of the two numbers gives 17.
* * *
The equations becomes:
\\[
12x^2 +20x -3x - 5 = 0
\\]
Factoring the common terms, we get: \\begin{align\*} & 12x^2 -3x +20x- 5 = 0 \\[0.6em] & 3x(4x-1) + 5(4x-1) = 0 \\[0.6em] & (3x + 5)(4x-1) = 0 \\end{align\*}
The solutions are the values of \\(x\\) for which \\(3x + 5 = 0\\) and \\(4x-1 = 0\\)
\\[
3x + 5 = 0 \\to 3x=-5 \\to x = -\\frac{5}{3}
\\]
\\[
4x-1 = 0 \\to 4x=1 \\to x = \\frac{1}{4}
\\]
The solution to the equation is:
\\[
x = -\\frac{5}{3} \\quad \\quad x = \\frac{1}{4}
\\]
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}
\\]
