2t^2+4t-3=0

Solution for 2t^2+4t-3=0 equation:

2t^2+4t-3=0

a = 2; b = 4; c = -3;
Δ = b2-4ac
Δ = 42-4·2·(-3)
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{10}}{2*2}=\frac{-4-2\sqrt{10}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{10}}{2*2}=\frac{-4+2\sqrt{10}}{4}
$