21r^2+12=32r

Solution for 21r^2+12=32r equation:

21r^2+12=32r

We move all terms to the left:

21r^2+12-(32r)=0

a = 21; b = -32; c = +12;
Δ = b2-4ac
Δ = -322-4·21·12
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4}{2*21}=\frac{28}{42}
=2/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4}{2*21}=\frac{36}{42}
=6/7
$