125x^2+49x-196=0

Solution for 125x^2+49x-196=0 equation:

125x^2+49x-196=0

a = 125; b = 49; c = -196;
Δ = b2-4ac
Δ = 492-4·125·(-196)
Δ = 100401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{100401}=\sqrt{49*2049}=\sqrt{49}*\sqrt{2049}=7\sqrt{2049}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-7\sqrt{2049}}{2*125}=\frac{-49-7\sqrt{2049}}{250}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+7\sqrt{2049}}{2*125}=\frac{-49+7\sqrt{2049}}{250}
$