Question

Solve the equation: ${(-x+3)}^2-12=8$ Sum of roots is

Answer 1

Solve $(-x+3{)}^2-12=8$ as follows:

$x^2+9-6x-12=8$

$x^2-6x-3=8$

$x^2-6x=11$

$x^2-6x-11=0$

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

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

$=\frac{-(-6)\pm \sqrt{{\left(6\right)}^2-4\left(1\right)(-11)}}{2\left(1\right)}$

$=\frac{6\pm \sqrt{36+44}}{2}$

$=\frac{6\pm \sqrt{80}}{2}$

$=\frac{6\pm 4\sqrt{5}}{2}$

$=3\pm 2\sqrt{5}$

Therefore, $x=3\pm 2\sqrt{5}$ .