18t^2-50=0

Solution for 18t^2-50=0 equation:

18t^2-50=0

a = 18; b = 0; c = -50;
Δ = b2-4ac
Δ = 02-4·18·(-50)
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-60}{2*18}=\frac{-60}{36}
=-1+2/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+60}{2*18}=\frac{60}{36}
=1+2/3
$