t=-8(2t^2-9)

Solution for t=-8(2t^2-9) equation:

t=-8(2t^2-9)

We move all terms to the left:

t-(-8(2t^2-9))=0


We calculate terms in parentheses: -(-8(2t^2-9)), so:

-8(2t^2-9)

We multiply parentheses

-16t^2+72

Back to the equation:

-(-16t^2+72)


We get rid of parentheses

16t^2+t-72=0

a = 16; b = 1; c = -72;
Δ = b2-4ac
Δ = 12-4·16·(-72)
Δ = 4609
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{4609}}{2*16}=\frac{-1-\sqrt{4609}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{4609}}{2*16}=\frac{-1+\sqrt{4609}}{32}
$