20r^2-31+12=0

Solution for 20r^2-31+12=0 equation:

20r^2-31+12=0

We add all the numbers together, and all the variables

20r^2-19=0

a = 20; b = 0; c = -19;
Δ = b2-4ac
Δ = 02-4·20·(-19)
Δ = 1520
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:

$\sqrt{\Delta}=\sqrt{1520}=\sqrt{16*95}=\sqrt{16}*\sqrt{95}=4\sqrt{95}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{95}}{2*20}=\frac{0-4\sqrt{95}}{40}
=-\frac{4\sqrt{95}}{40}
=-\frac{\sqrt{95}}{10}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{95}}{2*20}=\frac{0+4\sqrt{95}}{40}
=\frac{4\sqrt{95}}{40}
=\frac{\sqrt{95}}{10}
$