21r^2+48r+12=0

Solution for 21r^2+48r+12=0 equation:

21r^2+48r+12=0

a = 21; b = 48; c = +12;
Δ = b2-4ac
Δ = 482-4·21·12
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-36}{2*21}=\frac{-84}{42}
=-2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+36}{2*21}=\frac{-12}{42}
=-2/7
$