7=-16t2+54t+40

Solution for 7=-16t2+54t+40 equation:

7=-16t^2+54t+40

We move all terms to the left:

7-(-16t^2+54t+40)=0

We get rid of parentheses

16t^2-54t-40+7=0

We add all the numbers together, and all the variables

16t^2-54t-33=0

a = 16; b = -54; c = -33;
Δ = b2-4ac
Δ = -542-4·16·(-33)
Δ = 5028
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:

$\sqrt{\Delta}=\sqrt{5028}=\sqrt{4*1257}=\sqrt{4}*\sqrt{1257}=2\sqrt{1257}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-54)-2\sqrt{1257}}{2*16}=\frac{54-2\sqrt{1257}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-54)+2\sqrt{1257}}{2*16}=\frac{54+2\sqrt{1257}}{32}
$