Please help (algebra)

$2x^2=4x+9$

$2x^2-9=4x$

$2x^2-4x-9=0$

$x_{1,\:2}=\frac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\cdot \:2\left(-9\right)}}{2\cdot \:2}$

$x_{1,\:2}=\frac{-\left(-4\right)\pm \:2\sqrt{22}}{2\cdot \:2}$

$x_1=\frac{-\left(-4\right)+2\sqrt{22}}{2\cdot \:2},\:x_2=\frac{-\left(-4\right)-2\sqrt{22}}{2\cdot \:2}$

$x=\frac{2+\sqrt{22}}{2},\:x=\frac{2-\sqrt{22}}{2}$

In this case, it is

$x=\frac{2+\sqrt{22}}{2}$

2+22+2=26

so 26 is your answer!