40x^2+12x+8=73

Solution for 40x^2+12x+8=73 equation:

40x^2+12x+8=73

We move all terms to the left:

40x^2+12x+8-(73)=0

We add all the numbers together, and all the variables

40x^2+12x-65=0

a = 40; b = 12; c = -65;
Δ = b2-4ac
Δ = 122-4·40·(-65)
Δ = 10544
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:

$\sqrt{\Delta}=\sqrt{10544}=\sqrt{16*659}=\sqrt{16}*\sqrt{659}=4\sqrt{659}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{659}}{2*40}=\frac{-12-4\sqrt{659}}{80}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{659}}{2*40}=\frac{-12+4\sqrt{659}}{80}
$