10=-16t^2+400

Solution for 10=-16t^2+400 equation:

10=-16t^2+400

We move all terms to the left:

10-(-16t^2+400)=0

We get rid of parentheses

16t^2-400+10=0

We add all the numbers together, and all the variables

16t^2-390=0

a = 16; b = 0; c = -390;
Δ = b2-4ac
Δ = 02-4·16·(-390)
Δ = 24960
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:

$\sqrt{\Delta}=\sqrt{24960}=\sqrt{64*390}=\sqrt{64}*\sqrt{390}=8\sqrt{390}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{390}}{2*16}=\frac{0-8\sqrt{390}}{32}
=-\frac{8\sqrt{390}}{32}
=-\frac{\sqrt{390}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{390}}{2*16}=\frac{0+8\sqrt{390}}{32}
=\frac{8\sqrt{390}}{32}
=\frac{\sqrt{390}}{4}
$