Question

Solve 2x2 − 4x − 5 = 0 by completing the square.

Answer 1

Rewrite the right side of the equation
$2x^2-4x-5 = 0$ in the following way:

$2x^2-4x-5 = 2(x^2-2x)-5=2(x^2-2x+1-1)-5=2((x-1)^2-1)-5=2(x-1)^2-2-5=2(x-2)^2-7$.
Then the equation is $2(x-2)^2-7=0$ and $2(x-2)^2=7$.

So, $(x-2)^2= \frac{7}{2}$ and $x-2=\pm \sqrt{ \frac{7}{2} }$
The equation has two solutions:$x_1=2+ \sqrt{ \frac{7}{2} }$ and $x_2=2- \sqrt{ \frac{7}{2} }$.

Answer 2

2(x^2-2x+1-1)+5=0
2(x-1)^2+3=0
(x-1)^2=-3/2
x-1=+-√-3÷2

\[x=1+√$\frac{-3}{2}$\]
and \[x=1-√-3/2\]