34x+80x^2=1

Solution for 34x+80x^2=1 equation:

34x+80x^2=1

We move all terms to the left:

34x+80x^2-(1)=0

a = 80; b = 34; c = -1;
Δ = b2-4ac
Δ = 342-4·80·(-1)
Δ = 1476
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:

$\sqrt{\Delta}=\sqrt{1476}=\sqrt{36*41}=\sqrt{36}*\sqrt{41}=6\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-6\sqrt{41}}{2*80}=\frac{-34-6\sqrt{41}}{160}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+6\sqrt{41}}{2*80}=\frac{-34+6\sqrt{41}}{160}
$