5r^2-16+12=0

Solution for 5r^2-16+12=0 equation:

5r^2-16+12=0

We add all the numbers together, and all the variables

5r^2-4=0

a = 5; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·5·(-4)
Δ = 80
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*5}=\frac{0-4\sqrt{5}}{10}
=-\frac{4\sqrt{5}}{10}
=-\frac{2\sqrt{5}}{5}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*5}=\frac{0+4\sqrt{5}}{10}
=\frac{4\sqrt{5}}{10}
=\frac{2\sqrt{5}}{5}
$