3r^2+8r=16

Solution for 3r^2+8r=16 equation:

3r^2+8r=16

We move all terms to the left:

3r^2+8r-(16)=0

a = 3; b = 8; c = -16;
Δ = b2-4ac
Δ = 82-4·3·(-16)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-16}{2*3}=\frac{-24}{6}
=-4
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16}{2*3}=\frac{8}{6}
=1+1/3
$