0=2t^2+17t+8

Solution for 0=2t^2+17t+8 equation:

0=2t^2+17t+8

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-2t^2-17t-8=0

a = -2; b = -17; c = -8;
Δ = b2-4ac
Δ = -172-4·(-2)·(-8)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-15}{2*-2}=\frac{2}{-4}
=-1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+15}{2*-2}=\frac{32}{-4}
=-8
$