25x^2+40x=15

Solution for 25x^2+40x=15 equation:

25x^2+40x=15

We move all terms to the left:

25x^2+40x-(15)=0

a = 25; b = 40; c = -15;
Δ = b2-4ac
Δ = 402-4·25·(-15)
Δ = 3100
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:

$\sqrt{\Delta}=\sqrt{3100}=\sqrt{100*31}=\sqrt{100}*\sqrt{31}=10\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-10\sqrt{31}}{2*25}=\frac{-40-10\sqrt{31}}{50}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{31}}{2*25}=\frac{-40+10\sqrt{31}}{50}
$