2t^2-3.1t-6.2=0

Solution for 2t^2-3.1t-6.2=0 equation:

2t^2-3.1t-6.2=0

a = 2; b = -3.1; c = -6.2;
Δ = b2-4ac
Δ = -3.12-4·2·(-6.2)
Δ = 59.21
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{-(-3.1)-\sqrt{59.21}}{2*2}=\frac{3.1-\sqrt{59.21}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3.1)+\sqrt{59.21}}{2*2}=\frac{3.1+\sqrt{59.21}}{4}
$