3t^2+18t-21=0

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

3t^2+18t-21=0

a = 3; b = 18; c = -21;
Δ = b2-4ac
Δ = 182-4·3·(-21)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-24}{2*3}=\frac{-42}{6}
=-7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+24}{2*3}=\frac{6}{6}
=1
$