3x2+15x-21=S0

Solution for 3x2+15x-21=S0 equation:

3x^2+15x-21=0

a = 3; b = 15; c = -21;
Δ = b2-4ac
Δ = 152-4·3·(-21)
Δ = 477
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{477}=\sqrt{9*53}=\sqrt{9}*\sqrt{53}=3\sqrt{53}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-3\sqrt{53}}{2*3}=\frac{-15-3\sqrt{53}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+3\sqrt{53}}{2*3}=\frac{-15+3\sqrt{53}}{6}
$