The root of the equation $2(1+i)x^2-4(2-i)x-5-3i=0$, where $i=\sqrt{-1}$, which has eater modulus is

Let the roots of the quadratic equation 2(1+i)x^2 - 4(2-i)x - 5 - 3i = 0 be found using the quadratic formula. First, we calculate the discriminant D = b^2 - 4ac, where a = 2+2i, b = -8+4i, and c = -5-3i, which results in D = 6-8i. Applying the square root to the discriminant gives an imaginary root of approximately -1.41 + 2.82i. Plugging this back into the quadratic formula yields a root whose real and imaginary parts evaluate to (3-5i)/2, which possesses the greater modulus.

• Thanks
• Wrong explanation