56=-16t^2+62+4

Solution for 56=-16t^2+62+4 equation:

56=-16t^2+62+4

We move all terms to the left:

56-(-16t^2+62+4)=0

We get rid of parentheses

16t^2-62-4+56=0

We add all the numbers together, and all the variables

16t^2-10=0

a = 16; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·16·(-10)
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{10}}{2*16}=\frac{0-8\sqrt{10}}{32}
=-\frac{8\sqrt{10}}{32}
=-\frac{\sqrt{10}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{10}}{2*16}=\frac{0+8\sqrt{10}}{32}
=\frac{8\sqrt{10}}{32}
=\frac{\sqrt{10}}{4}
$