25x2+40x+2=0

Solution for 25x2+40x+2=0 equation:

25x^2+40x+2=0

a = 25; b = 40; c = +2;
Δ = b2-4ac
Δ = 402-4·25·2
Δ = 1400
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:

$\sqrt{\Delta}=\sqrt{1400}=\sqrt{100*14}=\sqrt{100}*\sqrt{14}=10\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-10\sqrt{14}}{2*25}=\frac{-40-10\sqrt{14}}{50}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{14}}{2*25}=\frac{-40+10\sqrt{14}}{50}
$