0=(-16t^2)-4t+472

Solution for 0=(-16t^2)-4t+472 equation:

0=(-16t^2)-4t+472

We move all terms to the left:

0-((-16t^2)-4t+472)=0

We add all the numbers together, and all the variables

-((-16t^2)-4t+472)=0


We calculate terms in parentheses: -((-16t^2)-4t+472), so:

(-16t^2)-4t+472

We get rid of parentheses

-16t^2-4t+472

Back to the equation:

-(-16t^2-4t+472)


We get rid of parentheses

16t^2+4t-472=0

a = 16; b = 4; c = -472;
Δ = b2-4ac
Δ = 42-4·16·(-472)
Δ = 30224
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:

$\sqrt{\Delta}=\sqrt{30224}=\sqrt{16*1889}=\sqrt{16}*\sqrt{1889}=4\sqrt{1889}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{1889}}{2*16}=\frac{-4-4\sqrt{1889}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{1889}}{2*16}=\frac{-4+4\sqrt{1889}}{32}
$