9x^2-40x-5=06

Solution for 9x^2-40x-5=06 equation:

9x^2-40x-5=06

We move all terms to the left:

9x^2-40x-5-(06)=0

We add all the numbers together, and all the variables

9x^2-40x-11=0

a = 9; b = -40; c = -11;
Δ = b2-4ac
Δ = -402-4·9·(-11)
Δ = 1996
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1996}=\sqrt{4*499}=\sqrt{4}*\sqrt{499}=2\sqrt{499}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-2\sqrt{499}}{2*9}=\frac{40-2\sqrt{499}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+2\sqrt{499}}{2*9}=\frac{40+2\sqrt{499}}{18}
$