2t^2+6t-3=0

Solution for 2t^2+6t-3=0 equation:

2t^2+6t-3=0

a = 2; b = 6; c = -3;
Δ = b2-4ac
Δ = 62-4·2·(-3)
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{15}}{2*2}=\frac{-6-2\sqrt{15}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{15}}{2*2}=\frac{-6+2\sqrt{15}}{4}
$