-5t^2+30=0

Solution for -5t^2+30=0 equation:

-5t^2+30=0

a = -5; b = 0; c = +30;
Δ = b2-4ac
Δ = 02-4·(-5)·30
Δ = 600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{600}=\sqrt{100*6}=\sqrt{100}*\sqrt{6}=10\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{6}}{2*-5}=\frac{0-10\sqrt{6}}{-10}
=-\frac{10\sqrt{6}}{-10}
=-\frac{\sqrt{6}}{-1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{6}}{2*-5}=\frac{0+10\sqrt{6}}{-10}
=\frac{10\sqrt{6}}{-10}
=\frac{\sqrt{6}}{-1}
$