-15x^2+256=0

Solution for -15x^2+256=0 equation:

-15x^2+256=0

a = -15; b = 0; c = +256;
Δ = b2-4ac
Δ = 02-4·(-15)·256
Δ = 15360
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:

$\sqrt{\Delta}=\sqrt{15360}=\sqrt{1024*15}=\sqrt{1024}*\sqrt{15}=32\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{15}}{2*-15}=\frac{0-32\sqrt{15}}{-30}
=-\frac{32\sqrt{15}}{-30}
=-\frac{16\sqrt{15}}{-15}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{15}}{2*-15}=\frac{0+32\sqrt{15}}{-30}
=\frac{32\sqrt{15}}{-30}
=\frac{16\sqrt{15}}{-15}
$