(3)=-16t^2+70t+5

Solution for (3)=-16t^2+70t+5 equation:

(3)=-16t^2+70t+5

We move all terms to the left:

(3)-(-16t^2+70t+5)=0

We get rid of parentheses

16t^2-70t-5+3=0

We add all the numbers together, and all the variables

16t^2-70t-2=0

a = 16; b = -70; c = -2;
Δ = b2-4ac
Δ = -702-4·16·(-2)
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{-(-70)-2\sqrt{1257}}{2*16}=\frac{70-2\sqrt{1257}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-70)+2\sqrt{1257}}{2*16}=\frac{70+2\sqrt{1257}}{32}
$