-23=-16(t^2-t)

Solution for -23=-16(t^2-t) equation:

-23=-16(t^2-t)

We move all terms to the left:

-23-(-16(t^2-t))=0


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

-16(t^2-t)

We multiply parentheses

-16t^2+16t

Back to the equation:

-(-16t^2+16t)


We get rid of parentheses

16t^2-16t-23=0

a = 16; b = -16; c = -23;
Δ = b2-4ac
Δ = -162-4·16·(-23)
Δ = 1728
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:

$\sqrt{\Delta}=\sqrt{1728}=\sqrt{576*3}=\sqrt{576}*\sqrt{3}=24\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-24\sqrt{3}}{2*16}=\frac{16-24\sqrt{3}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+24\sqrt{3}}{2*16}=\frac{16+24\sqrt{3}}{32}
$