Question

Find the number of real solutions to the equation
$2{(x-2)}^2-6=-2$

Answer 1

Solve the given equation as follows:

$2(x-2{)}^2-6=-2$

$2(x^2+4-4x)-6=-2$

$2x^2+8-8x-6=-2$

$2x^2-8x+4=0$

$x^2-4x+2=0$

Now we find the roots of above quadratic equation as shown below:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{-(-4)\pm \sqrt{{(-4)}^2-4\left(1\right)\left(2\right)}}{2\left(1\right)}$

$=\frac{4\pm \sqrt{16-8}}{2}$

$=\frac{4\pm \sqrt{8}}{2}$

$=\frac{4\pm 2\sqrt{2}}{2}$

$=2\pm \sqrt{2}$

Therefore, there are $2$ real solutions.