39x^2+500x-2500=0

Solution for 39x^2+500x-2500=0 equation:

39x^2+500x-2500=0

a = 39; b = 500; c = -2500;
Δ = b2-4ac
Δ = 5002-4·39·(-2500)
Δ = 640000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{640000}=800$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(500)-800}{2*39}=\frac{-1300}{78}
=-16+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(500)+800}{2*39}=\frac{300}{78}
=3+11/13
$