2t^2+6t-5=0

Solution for 2t^2+6t-5=0 equation:

2t^2+6t-5=0

a = 2; b = 6; c = -5;
Δ = b2-4ac
Δ = 62-4·2·(-5)
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{19}}{2*2}=\frac{-6-2\sqrt{19}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{19}}{2*2}=\frac{-6+2\sqrt{19}}{4}
$