3t^2+4t-3=0

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

3t^2+4t-3=0

a = 3; b = 4; c = -3;
Δ = b2-4ac
Δ = 42-4·3·(-3)
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{13}}{2*3}=\frac{-4-2\sqrt{13}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{13}}{2*3}=\frac{-4+2\sqrt{13}}{6}
$