(t)=-16t^2+30t+8

Solution for (t)=-16t^2+30t+8 equation:

(t)=-16t^2+30t+8

We move all terms to the left:

(t)-(-16t^2+30t+8)=0

We get rid of parentheses

16t^2-30t+t-8=0

We add all the numbers together, and all the variables

16t^2-29t-8=0

a = 16; b = -29; c = -8;
Δ = b2-4ac
Δ = -292-4·16·(-8)
Δ = 1353
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{-(-29)-\sqrt{1353}}{2*16}=\frac{29-\sqrt{1353}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+\sqrt{1353}}{2*16}=\frac{29+\sqrt{1353}}{32}
$