0=-2(8t^2-14.5-3)

Solution for 0=-2(8t^2-14.5-3) equation:

0=-2(8t^2-14.5-3)

We move all terms to the left:

0-(-2(8t^2-14.5-3))=0

We add all the numbers together, and all the variables

-(-2(8t^2-14.5-3))=0


We calculate terms in parentheses: -(-2(8t^2-14.5-3)), so:

-2(8t^2-14.5-3)

We multiply parentheses

-16t^2+29+6

We add all the numbers together, and all the variables

-16t^2+35

Back to the equation:

-(-16t^2+35)


We get rid of parentheses

16t^2-35=0

a = 16; b = 0; c = -35;
Δ = b2-4ac
Δ = 02-4·16·(-35)
Δ = 2240
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:

$\sqrt{\Delta}=\sqrt{2240}=\sqrt{64*35}=\sqrt{64}*\sqrt{35}=8\sqrt{35}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{35}}{2*16}=\frac{0-8\sqrt{35}}{32}
=-\frac{8\sqrt{35}}{32}
=-\frac{\sqrt{35}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{35}}{2*16}=\frac{0+8\sqrt{35}}{32}
=\frac{8\sqrt{35}}{32}
=\frac{\sqrt{35}}{4}
$