Question

solve $-2x^2 - 5 = -2x$ using any method.

$x = \square$ and $x = \square$

Explanation:

Step1: Rewrite the equation in standard form

First, we want to rewrite the quadratic equation \(-2x^{2}-5 = -2x\) in the standard form \(ax^{2}+bx + c=0\). We can do this by adding \(2x\) to both sides of the equation:
\(-2x^{2}+2x - 5=0\)
We can also multiply both sides by \(- 1\) to make the coefficient of \(x^{2}\) positive:
\(2x^{2}-2x + 5=0\)
Here, \(a = 2\), \(b=-2\), and \(c = 5\).

Step2: Use the quadratic formula

The quadratic formula is given by \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, we calculate the discriminant \(\Delta=b^{2}-4ac\). Substitute \(a = 2\), \(b=-2\), and \(c = 5\) into the discriminant formula:
\(\Delta=(-2)^{2}-4\times2\times5\)
\(=4 - 40\)
\(=- 36\)

Since the discriminant \(\Delta=-36<0\), the solutions of the quadratic equation are complex numbers. We know that \(\sqrt{-36}=\sqrt{36}\times\sqrt{-1}=6i\) (where \(i=\sqrt{-1}\) is the imaginary unit).

Now, substitute \(a = 2\), \(b=-2\), and \(\Delta=-36\) into the quadratic formula:
\(x=\frac{-(-2)\pm\sqrt{-36}}{2\times2}=\frac{2\pm6i}{4}\)
Simplify the fraction:
\(x=\frac{1\pm3i}{2}\)
Which can be written as \(x=\frac{1 + 3i}{2}\) and \(x=\frac{1-3i}{2}\)

Answer:

\(x=\frac{1 + 3i}{2}\) and \(x=\frac{1-3i}{2}\)