1269-16t^2=0

Solution for 1269-16t^2=0 equation:

1269-16t^2=0

a = -16; b = 0; c = +1269;
Δ = b2-4ac
Δ = 02-4·(-16)·1269
Δ = 81216
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:

$\sqrt{\Delta}=\sqrt{81216}=\sqrt{576*141}=\sqrt{576}*\sqrt{141}=24\sqrt{141}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{141}}{2*-16}=\frac{0-24\sqrt{141}}{-32}
=-\frac{24\sqrt{141}}{-32}
=-\frac{3\sqrt{141}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{141}}{2*-16}=\frac{0+24\sqrt{141}}{-32}
=\frac{24\sqrt{141}}{-32}
=\frac{3\sqrt{141}}{-4}
$