16t^2+70t+6=-3

Solution for 16t^2+70t+6=-3 equation:

16t^2+70t+6=-3

We move all terms to the left:

16t^2+70t+6-(-3)=0

We add all the numbers together, and all the variables

16t^2+70t+9=0

a = 16; b = 70; c = +9;
Δ = b2-4ac
Δ = 702-4·16·9
Δ = 4324
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:

$\sqrt{\Delta}=\sqrt{4324}=\sqrt{4*1081}=\sqrt{4}*\sqrt{1081}=2\sqrt{1081}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(70)-2\sqrt{1081}}{2*16}=\frac{-70-2\sqrt{1081}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(70)+2\sqrt{1081}}{2*16}=\frac{-70+2\sqrt{1081}}{32}
$