400x^2+80x-5=0

Solution for 400x^2+80x-5=0 equation:

400x^2+80x-5=0

a = 400; b = 80; c = -5;
Δ = b2-4ac
Δ = 802-4·400·(-5)
Δ = 14400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{14400}=120$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-120}{2*400}=\frac{-200}{800}
=-1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+120}{2*400}=\frac{40}{800}
=1/20
$