25a^2+3a-122=0

Solution for 25a^2+3a-122=0 equation:

25a^2+3a-122=0

a = 25; b = 3; c = -122;
Δ = b2-4ac
Δ = 32-4·25·(-122)
Δ = 12209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{12209}}{2*25}=\frac{-3-\sqrt{12209}}{50}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{12209}}{2*25}=\frac{-3+\sqrt{12209}}{50}
$