p) \( 16 x^{2}+2 x+7=10 x-6 \)

To solve the equation \( 16 x^{2}+2 x+7=10 x-6 \), first, let's rearrange it by moving all terms to one side. This gives us: \( 16 x^{2} + 2 x + 7 - 10 x + 6 = 0 \) Simplifying this, we get: \( 16 x^{2} - 8 x + 13 = 0 \) Now, we can apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 16 \), \( b = -8 \), and \( c = 13 \). Plugging in the values yields: 1. Calculate \( b^2 - 4ac = (-8)^2 - 4 \cdot 16 \cdot 13 \). 2. This simplifies to \( 64 - 832 = -768 \). Since the discriminant is negative, this means there are no real solutions—only complex solutions exist! To find the complex roots, we continue with: \[ x = \frac{8 \pm \sqrt{-768}}{32} \] \[ x = \frac{8 \pm 16i\sqrt{3}}{32} \] \[ x = \frac{1 \pm 2i\sqrt{3}}{4} \] So the solutions are \( x = \frac{1 + 2i\sqrt{3}}{4} \) and \( x = \frac{1 - 2i\sqrt{3}}{4} \).