8r^2+8=40

Solution for 8r^2+8=40 equation:

8r^2+8=40

We move all terms to the left:

8r^2+8-(40)=0

We add all the numbers together, and all the variables

8r^2-32=0

a = 8; b = 0; c = -32;
Δ = b2-4ac
Δ = 02-4·8·(-32)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32}{2*8}=\frac{-32}{16}
=-2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32}{2*8}=\frac{32}{16}
=2
$