Question

Solve the following equations,$2(1+i)x^2-4(2-i)x-5-3i=0.$

Answer 1

$2(1+i)x^2-4(2-i)x-5-3i=0.$

or, $x=\frac{4(2-i)\pm \sqrt{16(2-i{)}^2+8(1+i\left)\right(5+3i)}}{4(1+i)}$

or, $x=\frac{4(2-i)\pm \sqrt{8\left\{2\right(4-4i-1)+(5-3+8i\left)\right\}}}{4(1+i)}$

or, $x=\frac{4(2-i)\pm \sqrt{8\times 8}}{4(1+i)}$

or, $x=\frac{4(2-i)\pm 8}{4(1+i)}$

or, $x=\frac{-4i}{4(1+i)}$ or $x=\frac{4(4-i)}{4(1+i)}$

or, $x=\frac{-i}{(1+i)}$ or $x=\frac{(4-i)}{(1+i)}$

or, $x=\frac{-i(1-i)}{\left(2\right)}$ or $x=\frac{(4-i)(1-i)}{\left(2\right)}$

or, $x=\frac{-(1+i)}{\left(2\right)}$ or $x=\frac{(3-5i)}{\left(2\right)}$