160t^2-490=0

Solution for 160t^2-490=0 equation:

160t^2-490=0

a = 160; b = 0; c = -490;
Δ = b2-4ac
Δ = 02-4·160·(-490)
Δ = 313600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{313600}=560$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-560}{2*160}=\frac{-560}{320}
=-1+3/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+560}{2*160}=\frac{560}{320}
=1+3/4
$