4r2+12r=112

Solution for 4r2+12r=112 equation:

4r^2+12r=112

We move all terms to the left:

4r^2+12r-(112)=0

a = 4; b = 12; c = -112;
Δ = b2-4ac
Δ = 122-4·4·(-112)
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-44}{2*4}=\frac{-56}{8}
=-7
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+44}{2*4}=\frac{32}{8}
=4
$