0=-16(t2-6t-7)

Solution for 0=-16(t2-6t-7) equation:

0=-16(t2-6t-7)

We move all terms to the left:

0-(-16(t2-6t-7))=0

We add all the numbers together, and all the variables

-(-16(+t^2-6t-7))+0=0

We add all the numbers together, and all the variables

-(-16(+t^2-6t-7))=0


We calculate terms in parentheses: -(-16(+t^2-6t-7)), so:

-16(+t^2-6t-7)

We multiply parentheses

-16t^2+96t+112

Back to the equation:

-(-16t^2+96t+112)


We get rid of parentheses

16t^2-96t-112=0

a = 16; b = -96; c = -112;
Δ = b2-4ac
Δ = -962-4·16·(-112)
Δ = 16384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16384}=128$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-128}{2*16}=\frac{-32}{32}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+128}{2*16}=\frac{224}{32}
=7
$