r2+16r+48=0

Solution for r2+16r+48=0 equation:

r2+16r+48=0

We add all the numbers together, and all the variables

r^2+16r+48=0

a = 1; b = 16; c = +48;
Δ = b2-4ac
Δ = 162-4·1·48
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8}{2*1}=\frac{-24}{2}
=-12
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8}{2*1}=\frac{-8}{2}
=-4
$