2a+5=a^2

Solution for 2a+5=a^2 equation:

2a+5=a^2

We move all terms to the left:

2a+5-(a^2)=0

determiningTheFunctionDomain
-a^2+2a+5=0

We add all the numbers together, and all the variables

-1a^2+2a+5=0

a = -1; b = 2; c = +5;
Δ = b2-4ac
Δ = 22-4·(-1)·5
Δ = 24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{6}}{2*-1}=\frac{-2-2\sqrt{6}}{-2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{6}}{2*-1}=\frac{-2+2\sqrt{6}}{-2}
$