Question 2 \( 2.12 x^{2}+2 x+1=0 \) solve for \( x \)

Let’s use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2.12 \), \( b = 2 \), and \( c = 1 \). Plugging in these values gives us \( x = \frac{-2 \pm \sqrt{(2)^2 - 4(2.12)(1)}}{2(2.12)} \). Simplifying, we find that the discriminant is negative, indicating there are no real solutions—only complex ones! So, we end up with \( x = \frac{-2 \pm \sqrt{-4.48}}{4.24} \), which can be expressed as \( x = \frac{-2 \pm i\sqrt{4.48}}{4.24} \). In essence, this equation opens the door to the fascinating world of complex numbers, a realm where the imaginary unit \( i \) (where \( i^2 = -1 \)) allows us to explore "solutions" we can't see on the typical real number line. So when life gives you a quadratic with no real solutions, grab that imaginary friend and dive in!