3t^2+8t-10=0

Solution for 3t^2+8t-10=0 equation:

3t^2+8t-10=0

a = 3; b = 8; c = -10;
Δ = b2-4ac
Δ = 82-4·3·(-10)
Δ = 184
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:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{46}}{2*3}=\frac{-8-2\sqrt{46}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{46}}{2*3}=\frac{-8+2\sqrt{46}}{6}
$