3r^2+8r-48=0

Solution for 3r^2+8r-48=0 equation:

3r^2+8r-48=0

a = 3; b = 8; c = -48;
Δ = b2-4ac
Δ = 82-4·3·(-48)
Δ = 640
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{10}}{2*3}=\frac{-8-8\sqrt{10}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{10}}{2*3}=\frac{-8+8\sqrt{10}}{6}
$