Question

Can you help me solve this radical equation √y-10 = y-2 the sqrt sign is only over the y-10

Answer 1

$D:y-10\geq0 \wedge y-2\geq0\\ D:y\geq10 \wedge y\geq2\\ D:y\geq10\\\\ \sqrt{y-10}=y-2\\ y-10=(y-2)^2\\ y-10=y^2-4y+4\\ y^2-5y+14=0\\ \Delta=(-5)^2-4\cdot1\cdot14=25-56=-31\\ \Delta$