25x2+50x+12=0

Solution for 25x2+50x+12=0 equation:

25x^2+50x+12=0

a = 25; b = 50; c = +12;
Δ = b2-4ac
Δ = 502-4·25·12
Δ = 1300
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:

$\sqrt{\Delta}=\sqrt{1300}=\sqrt{100*13}=\sqrt{100}*\sqrt{13}=10\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-10\sqrt{13}}{2*25}=\frac{-50-10\sqrt{13}}{50}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+10\sqrt{13}}{2*25}=\frac{-50+10\sqrt{13}}{50}
$