-5x^2+27=0

Solution for -5x^2+27=0 equation:

-5x^2+27=0

a = -5; b = 0; c = +27;
Δ = b2-4ac
Δ = 02-4·(-5)·27
Δ = 540
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:

$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{15}}{2*-5}=\frac{0-6\sqrt{15}}{-10}
=-\frac{6\sqrt{15}}{-10}
=-\frac{3\sqrt{15}}{-5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{15}}{2*-5}=\frac{0+6\sqrt{15}}{-10}
=\frac{6\sqrt{15}}{-10}
=\frac{3\sqrt{15}}{-5}
$