=16t^2-312

Solution for =16t^2-312 equation:

=16t^2-312

We move all terms to the left:

-(16t^2-312)=0

We get rid of parentheses

-16t^2+312=0

a = -16; b = 0; c = +312;
Δ = b2-4ac
Δ = 02-4·(-16)·312
Δ = 19968
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:

$\sqrt{\Delta}=\sqrt{19968}=\sqrt{256*78}=\sqrt{256}*\sqrt{78}=16\sqrt{78}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{78}}{2*-16}=\frac{0-16\sqrt{78}}{-32}
=-\frac{16\sqrt{78}}{-32}
=-\frac{\sqrt{78}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{78}}{2*-16}=\frac{0+16\sqrt{78}}{-32}
=\frac{16\sqrt{78}}{-32}
=\frac{\sqrt{78}}{-2}
$