25x^2+40x=-12

Solution for 25x^2+40x=-12 equation:

25x^2+40x=-12

We move all terms to the left:

25x^2+40x-(-12)=0

We add all the numbers together, and all the variables

25x^2+40x+12=0

a = 25; b = 40; c = +12;
Δ = b2-4ac
Δ = 402-4·25·12
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20}{2*25}=\frac{-60}{50}
=-1+1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20}{2*25}=\frac{-20}{50}
=-2/5
$