3t^2+6t-40=0

Solution for 3t^2+6t-40=0 equation:

3t^2+6t-40=0

a = 3; b = 6; c = -40;
Δ = b2-4ac
Δ = 62-4·3·(-40)
Δ = 516
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:

$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{129}}{2*3}=\frac{-6-2\sqrt{129}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{129}}{2*3}=\frac{-6+2\sqrt{129}}{6}
$