3r^2+16r-12=0

Solution for 3r^2+16r-12=0 equation:

3r^2+16r-12=0

a = 3; b = 16; c = -12;
Δ = b2-4ac
Δ = 162-4·3·(-12)
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{-(16)-20}{2*3}=\frac{-36}{6}
=-6
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+20}{2*3}=\frac{4}{6}
=2/3
$