(8t-4t^2)+17(6t+2)=0

Solution for (8t-4t^2)+17(6t+2)=0 equation:

(8t-4t^2)+17(6t+2)=0

We multiply parentheses

(8t-4t^2)+102t+34=0

We get rid of parentheses

-4t^2+8t+102t+34=0

We add all the numbers together, and all the variables

-4t^2+110t+34=0

a = -4; b = 110; c = +34;
Δ = b2-4ac
Δ = 1102-4·(-4)·34
Δ = 12644
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:

$\sqrt{\Delta}=\sqrt{12644}=\sqrt{4*3161}=\sqrt{4}*\sqrt{3161}=2\sqrt{3161}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(110)-2\sqrt{3161}}{2*-4}=\frac{-110-2\sqrt{3161}}{-8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(110)+2\sqrt{3161}}{2*-4}=\frac{-110+2\sqrt{3161}}{-8}
$