7r^2+12r+4=8

Solution for 7r^2+12r+4=8 equation:

7r^2+12r+4=8

We move all terms to the left:

7r^2+12r+4-(8)=0

We add all the numbers together, and all the variables

7r^2+12r-4=0

a = 7; b = 12; c = -4;
Δ = b2-4ac
Δ = 122-4·7·(-4)
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{-(12)-16}{2*7}=\frac{-28}{14}
=-2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+16}{2*7}=\frac{4}{14}
=2/7
$