2r^2-4r-48=0

Solution for 2r^2-4r-48=0 equation:

2r^2-4r-48=0

a = 2; b = -4; c = -48;
Δ = b2-4ac
Δ = -42-4·2·(-48)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-20}{2*2}=\frac{-16}{4}
=-4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+20}{2*2}=\frac{24}{4}
=6
$