10=6+17-16t^2

Solution for 10=6+17-16t^2 equation:

10=6+17-16t^2

We move all terms to the left:

10-(6+17-16t^2)=0

We get rid of parentheses

16t^2-6-17+10=0

We add all the numbers together, and all the variables

16t^2-13=0

a = 16; b = 0; c = -13;
Δ = b2-4ac
Δ = 02-4·16·(-13)
Δ = 832
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:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{13}}{2*16}=\frac{0-8\sqrt{13}}{32}
=-\frac{8\sqrt{13}}{32}
=-\frac{\sqrt{13}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{13}}{2*16}=\frac{0+8\sqrt{13}}{32}
=\frac{8\sqrt{13}}{32}
=\frac{\sqrt{13}}{4}
$