3t^2+21t+18=0

Solution for 3t^2+21t+18=0 equation:

3t^2+21t+18=0

a = 3; b = 21; c = +18;
Δ = b2-4ac
Δ = 212-4·3·18
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-15}{2*3}=\frac{-36}{6}
=-6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+15}{2*3}=\frac{-6}{6}
=-1
$