40x2+-13x+-1=0

Solution for 40x2+-13x+-1=0 equation:

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

We add all the numbers together, and all the variables

40x^2-13x=0

a = 40; b = -13; c = 0;
Δ = b2-4ac
Δ = -132-4·40·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-13}{2*40}=\frac{0}{80}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+13}{2*40}=\frac{26}{80}
=13/40
$