0=-(16t^2-38t-5)

Solution for 0=-(16t^2-38t-5) equation:

0=-(16t^2-38t-5)

We move all terms to the left:

0-(-(16t^2-38t-5))=0

We add all the numbers together, and all the variables

-(-(16t^2-38t-5))=0


We calculate terms in parentheses: -(-(16t^2-38t-5)), so:

-(16t^2-38t-5)

We get rid of parentheses

-16t^2+38t+5

Back to the equation:

-(-16t^2+38t+5)


We get rid of parentheses

16t^2-38t-5=0

a = 16; b = -38; c = -5;
Δ = b2-4ac
Δ = -382-4·16·(-5)
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-42}{2*16}=\frac{-4}{32}
=-1/8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+42}{2*16}=\frac{80}{32}
=2+1/2
$