2r^2+8r-48=0

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

2r^2+8r-48=0

a = 2; b = 8; c = -48;
Δ = b2-4ac
Δ = 82-4·2·(-48)
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{7}}{2*2}=\frac{-8-8\sqrt{7}}{4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{7}}{2*2}=\frac{-8+8\sqrt{7}}{4}
$