3s^2-22s-48=0

Solution for 3s^2-22s-48=0 equation:

3s^2-22s-48=0

a = 3; b = -22; c = -48;
Δ = b2-4ac
Δ = -222-4·3·(-48)
Δ = 1060
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1060}=\sqrt{4*265}=\sqrt{4}*\sqrt{265}=2\sqrt{265}$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{265}}{2*3}=\frac{22-2\sqrt{265}}{6}
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{265}}{2*3}=\frac{22+2\sqrt{265}}{6}
$