15t^2-4t-2=0

Solution for 15t^2-4t-2=0 equation:

15t^2-4t-2=0

a = 15; b = -4; c = -2;
Δ = b2-4ac
Δ = -42-4·15·(-2)
Δ = 136
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:

$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{34}}{2*15}=\frac{4-2\sqrt{34}}{30}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{34}}{2*15}=\frac{4+2\sqrt{34}}{30}
$