-25x^2+13=0

Solution for -25x^2+13=0 equation:

-25x^2+13=0

a = -25; b = 0; c = +13;
Δ = b2-4ac
Δ = 02-4·(-25)·13
Δ = 1300
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:

$\sqrt{\Delta}=\sqrt{1300}=\sqrt{100*13}=\sqrt{100}*\sqrt{13}=10\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{13}}{2*-25}=\frac{0-10\sqrt{13}}{-50}
=-\frac{10\sqrt{13}}{-50}
=-\frac{\sqrt{13}}{-5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{13}}{2*-25}=\frac{0+10\sqrt{13}}{-50}
=\frac{10\sqrt{13}}{-50}
=\frac{\sqrt{13}}{-5}
$