3t^2+22t+40=0

Solution for 3t^2+22t+40=0 equation:

3t^2+22t+40=0

a = 3; b = 22; c = +40;
Δ = b2-4ac
Δ = 222-4·3·40
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2}{2*3}=\frac{-24}{6}
=-4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2}{2*3}=\frac{-20}{6}
=-3+1/3
$