106=16(t)^2+32(t)

Solution for 106=16(t)^2+32(t) equation:

106=16(t)^2+32(t)

We move all terms to the left:

106-(16(t)^2+32(t))=0

We get rid of parentheses

-16t^2-32t+106=0

a = -16; b = -32; c = +106;
Δ = b2-4ac
Δ = -322-4·(-16)·106
Δ = 7808
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:

$\sqrt{\Delta}=\sqrt{7808}=\sqrt{64*122}=\sqrt{64}*\sqrt{122}=8\sqrt{122}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{122}}{2*-16}=\frac{32-8\sqrt{122}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{122}}{2*-16}=\frac{32+8\sqrt{122}}{-32}
$