25x^2+25x-5=0

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

25x^2+25x-5=0

a = 25; b = 25; c = -5;
Δ = b2-4ac
Δ = 252-4·25·(-5)
Δ = 1125
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:

$\sqrt{\Delta}=\sqrt{1125}=\sqrt{225*5}=\sqrt{225}*\sqrt{5}=15\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-15\sqrt{5}}{2*25}=\frac{-25-15\sqrt{5}}{50}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+15\sqrt{5}}{2*25}=\frac{-25+15\sqrt{5}}{50}
$