-r^2+5r^2-4=0

Solution for -r^2+5r^2-4=0 equation:

-r^2+5r^2-4=0

We add all the numbers together, and all the variables

4r^2-4=0

a = 4; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·4·(-4)
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{-(0)-8}{2*4}=\frac{-8}{8}
=-1
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*4}=\frac{8}{8}
=1
$