Question

Find the number of real solutions to the quadratic equation
$x^2+2=x+5$

Answer 1

Solve $x^2+2=x+5$ as follows:

$x^2+2-x-5=0$

$x^2-x-3=0$

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

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

$=\frac{-(-1)\pm \sqrt{{(-1)}^2-4\left(1\right)(-3)}}{2\left(1\right)}$

$=\frac{1\pm \sqrt{1+12}}{2}$

$=\frac{1\pm \sqrt{13}}{2}$

Therefore,$x=\frac{1\pm \sqrt{13}}{2}$

Hence, the quadratic equation has $2$ real solutions