# Quadratic Equation A3

Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/exercises/quadratic-equation-a-3/
Fetched from algebrica.org test 3458; source modified 2025-03-06T16:14:07.

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

\\[
-7x + 3 = -2x^2
\\]

* * *

First, we need to rewrite the second-degree equation in its standard form \\(ax^2+bx+c = 0\\). By collecting all terms on the left side of the equal sign, we obtain::

\\[
2x^2-7x + 3 = 0
\\]

* * *

After reducing the equation to its standard form, we can substitute the coefficients \\(a=2, b=-7, c=3\\) into the [quadratic formula](../quadratic-formula.md):

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

We obtain:

\\[
x\_{1,2} = \\frac{{-(-7) \\pm \\sqrt{{(-7)^2-4(2)(3)}}}}{{2(2)}}
\\]

* * *

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

\\begin{align\*} x\_{1,2} &= \\frac{{7 \\pm \\sqrt{{49-24}}}}{4}\\[0.6em] &= \\frac{{7 \\pm \\sqrt{{25}}}}{4}\\[0.6em] &= \\frac{7 \\pm 5}{4}\\[0.6em] \\end{align\*}

* * *

Finally, by performing the calculations, we obtain:

\\[
x\_1=\\frac{7 + 5}{4}=\\frac{12}{4} = 3
\\]



\\[
x\_2=\\frac{7 - 5}{4}=\\frac{2}{4} = \\frac{1}{2}
\\]

The solution to the equation is:

\\[
x\_1 = 3 \\quad \\quad x\_2=\\frac{1}{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}
\\]