25x^2+40x=13

Solution for 25x^2+40x=13 equation:

25x^2+40x=13

We move all terms to the left:

25x^2+40x-(13)=0

a = 25; b = 40; c = -13;
Δ = b2-4ac
Δ = 402-4·25·(-13)
Δ = 2900
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:

$\sqrt{\Delta}=\sqrt{2900}=\sqrt{100*29}=\sqrt{100}*\sqrt{29}=10\sqrt{29}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-10\sqrt{29}}{2*25}=\frac{-40-10\sqrt{29}}{50}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{29}}{2*25}=\frac{-40+10\sqrt{29}}{50}
$