3x^2+21x-1620=0

Solution for 3x^2+21x-1620=0 equation:

3x^2+21x-1620=0

a = 3; b = 21; c = -1620;
Δ = b2-4ac
Δ = 212-4·3·(-1620)
Δ = 19881
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{19881}=141$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-141}{2*3}=\frac{-162}{6}
=-27
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+141}{2*3}=\frac{120}{6}
=20
$