25x^2+10x-5=0

Solution for 25x^2+10x-5=0 equation:

25x^2+10x-5=0

a = 25; b = 10; c = -5;
Δ = b2-4ac
Δ = 102-4·25·(-5)
Δ = 600
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:

$\sqrt{\Delta}=\sqrt{600}=\sqrt{100*6}=\sqrt{100}*\sqrt{6}=10\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{6}}{2*25}=\frac{-10-10\sqrt{6}}{50}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{6}}{2*25}=\frac{-10+10\sqrt{6}}{50}
$