3t^2+19t+30=0

Solution for 3t^2+19t+30=0 equation:

3t^2+19t+30=0

a = 3; b = 19; c = +30;
Δ = b2-4ac
Δ = 192-4·3·30
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-1}{2*3}=\frac{-20}{6}
=-3+1/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+1}{2*3}=\frac{-18}{6}
=-3
$