40x^2+13x=1

Solution for 40x^2+13x=1 equation:

40x^2+13x=1

We move all terms to the left:

40x^2+13x-(1)=0

a = 40; b = 13; c = -1;
Δ = b2-4ac
Δ = 132-4·40·(-1)
Δ = 329
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{329}}{2*40}=\frac{-13-\sqrt{329}}{80}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{329}}{2*40}=\frac{-13+\sqrt{329}}{80}
$