16t2+400t=-4

Solution for 16t2+400t=-4 equation:

16t^2+400t=-4

We move all terms to the left:

16t^2+400t-(-4)=0

We add all the numbers together, and all the variables

16t^2+400t+4=0

a = 16; b = 400; c = +4;
Δ = b2-4ac
Δ = 4002-4·16·4
Δ = 159744
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:

$\sqrt{\Delta}=\sqrt{159744}=\sqrt{4096*39}=\sqrt{4096}*\sqrt{39}=64\sqrt{39}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(400)-64\sqrt{39}}{2*16}=\frac{-400-64\sqrt{39}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(400)+64\sqrt{39}}{2*16}=\frac{-400+64\sqrt{39}}{32}
$