25x^2-16x-41=0

Solution for 25x^2-16x-41=0 equation:

25x^2-16x-41=0

a = 25; b = -16; c = -41;
Δ = b2-4ac
Δ = -162-4·25·(-41)
Δ = 4356
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4356}=66$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-66}{2*25}=\frac{-50}{50}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+66}{2*25}=\frac{82}{50}
=1+16/25
$