Question

5x² + 28 - 12 = 0

Explanation:

Step1: Simplify the equation

The given equation is \(5x^{2}+28 - 12=0\). First, simplify the constant terms: \(28-12 = 16\), so the equation becomes \(5x^{2}+16 = 0\). Then, rearrange it to solve for \(x^{2}\): \(5x^{2}=- 16\), so \(x^{2}=-\frac{16}{5}\).

Step2: Solve for x

Since the square of a real number cannot be negative, we consider the complex number system. Using the formula \(x=\pm\sqrt{a}\) when \(x^{2}=a\), here \(a =-\frac{16}{5}\), and we know that \(\sqrt{-1}=i\), so \(x=\pm\sqrt{\frac{16}{5}}\times i=\pm\frac{4}{\sqrt{5}}i\). Rationalizing the denominator, we get \(x=\pm\frac{4\sqrt{5}}{5}i\).

Answer:

The solutions of the equation \(5x^{2}+28 - 12 = 0\) are \(x=\frac{4\sqrt{5}}{5}i\) and \(x =-\frac{4\sqrt{5}}{5}i\) (or in boxed form, if we consider the complex solutions: \(\boxed{\pm\frac{4\sqrt{5}}{5}i}\))