a2-9=40a=

Solution for a2-9=40a= equation:

a2-9=40a=

We move all terms to the left:

a2-9-(40a)=0

We add all the numbers together, and all the variables

a^2-40a-9=0

a = 1; b = -40; c = -9;
Δ = b2-4ac
Δ = -402-4·1·(-9)
Δ = 1636
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:

$\sqrt{\Delta}=\sqrt{1636}=\sqrt{4*409}=\sqrt{4}*\sqrt{409}=2\sqrt{409}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-2\sqrt{409}}{2*1}=\frac{40-2\sqrt{409}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+2\sqrt{409}}{2*1}=\frac{40+2\sqrt{409}}{2}
$