-x^2+13=0

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

-x^2+13=0

We add all the numbers together, and all the variables

-1x^2+13=0

a = -1; b = 0; c = +13;
Δ = b2-4ac
Δ = 02-4·(-1)·13
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{13}}{2*-1}=\frac{0-2\sqrt{13}}{-2}
=-\frac{2\sqrt{13}}{-2}
=-\frac{\sqrt{13}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{13}}{2*-1}=\frac{0+2\sqrt{13}}{-2}
=\frac{2\sqrt{13}}{-2}
=\frac{\sqrt{13}}{-1}
$