-9t2+50=0

Solution for -9t2+50=0 equation:

-9t^2+50=0

a = -9; b = 0; c = +50;
Δ = b2-4ac
Δ = 02-4·(-9)·50
Δ = 1800
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:

$\sqrt{\Delta}=\sqrt{1800}=\sqrt{900*2}=\sqrt{900}*\sqrt{2}=30\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-30\sqrt{2}}{2*-9}=\frac{0-30\sqrt{2}}{-18}
=-\frac{30\sqrt{2}}{-18}
=-\frac{5\sqrt{2}}{-3}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+30\sqrt{2}}{2*-9}=\frac{0+30\sqrt{2}}{-18}
=\frac{30\sqrt{2}}{-18}
=\frac{5\sqrt{2}}{-3}
$