0=-16(t^2-8t-7)

Solution for 0=-16(t^2-8t-7) equation:

0=-16(t^2-8t-7)

We move all terms to the left:

0-(-16(t^2-8t-7))=0

We add all the numbers together, and all the variables

-(-16(t^2-8t-7))=0


We calculate terms in parentheses: -(-16(t^2-8t-7)), so:

-16(t^2-8t-7)

We multiply parentheses

-16t^2+128t+112

Back to the equation:

-(-16t^2+128t+112)


We get rid of parentheses

16t^2-128t-112=0

a = 16; b = -128; c = -112;
Δ = b2-4ac
Δ = -1282-4·16·(-112)
Δ = 23552
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:

$\sqrt{\Delta}=\sqrt{23552}=\sqrt{1024*23}=\sqrt{1024}*\sqrt{23}=32\sqrt{23}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-128)-32\sqrt{23}}{2*16}=\frac{128-32\sqrt{23}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-128)+32\sqrt{23}}{2*16}=\frac{128+32\sqrt{23}}{32}
$