-2t(t-19)=16

Solution for -2t(t-19)=16 equation:

-2t(t-19)=16

We move all terms to the left:

-2t(t-19)-(16)=0

We multiply parentheses

-2t^2+38t-16=0

a = -2; b = 38; c = -16;
Δ = b2-4ac
Δ = 382-4·(-2)·(-16)
Δ = 1316
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:

$\sqrt{\Delta}=\sqrt{1316}=\sqrt{4*329}=\sqrt{4}*\sqrt{329}=2\sqrt{329}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-2\sqrt{329}}{2*-2}=\frac{-38-2\sqrt{329}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+2\sqrt{329}}{2*-2}=\frac{-38+2\sqrt{329}}{-4}
$