Ref.: module A

This exercise requires understanding the fundamental concepts of quadratic equations and their solution methods. It involves using the quadratic formula

Solve the quadratic equation

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


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:

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

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