2.5t^2+12t-8=0

Solution for 2.5t^2+12t-8=0 equation:

2.5t^2+12t-8=0

a = 2.5; b = 12; c = -8;
Δ = b2-4ac
Δ = 122-4·2.5·(-8)
Δ = 224
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:

$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{14}}{2*2.5}=\frac{-12-4\sqrt{14}}{5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{14}}{2*2.5}=\frac{-12+4\sqrt{14}}{5}
$