16t^2-38+21=0

Solution for 16t^2-38+21=0 equation:

16t^2-38+21=0

We add all the numbers together, and all the variables

16t^2-17=0

a = 16; b = 0; c = -17;
Δ = b2-4ac
Δ = 02-4·16·(-17)
Δ = 1088
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:

$\sqrt{\Delta}=\sqrt{1088}=\sqrt{64*17}=\sqrt{64}*\sqrt{17}=8\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{17}}{2*16}=\frac{0-8\sqrt{17}}{32}
=-\frac{8\sqrt{17}}{32}
=-\frac{\sqrt{17}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{17}}{2*16}=\frac{0+8\sqrt{17}}{32}
=\frac{8\sqrt{17}}{32}
=\frac{\sqrt{17}}{4}
$