2t^2+1t-8=0

Solution for 2t^2+1t-8=0 equation:

2t^2+1t-8=0

We add all the numbers together, and all the variables

2t^2+t-8=0

a = 2; b = 1; c = -8;
Δ = b2-4ac
Δ = 12-4·2·(-8)
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{65}}{2*2}=\frac{-1-\sqrt{65}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{65}}{2*2}=\frac{-1+\sqrt{65}}{4}
$