give me a Quartic equation with 1 real solution?

HA! The trick is use the multiplicities! Here's an example:

x^4 - 4x³ + 8x² - 16x + 16 = 0 has only one real solution, x = 2, and two complex solutions x = ±2i

If the coefficients are all real numbers, it can't be done. A quadratic equation will always have four roots, including multiplicities. Complex roots always occur in conjugate pairs {a + bi and a - bi}, so there will always be 0, 2, or 4 complex roots. Hence there will always be 4, 2, or 0 real roots.