292=2r^2+4

Solution for 292=2r^2+4 equation:

292=2r^2+4

We move all terms to the left:

292-(2r^2+4)=0

We get rid of parentheses

-2r^2-4+292=0

We add all the numbers together, and all the variables

-2r^2+288=0

a = -2; b = 0; c = +288;
Δ = b2-4ac
Δ = 02-4·(-2)·288
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*-2}=\frac{-48}{-4}
=+12
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*-2}=\frac{48}{-4}
=-12
$