9m^2=3m+20

Solution for 9m^2=3m+20 equation:

9m^2=3m+20

We move all terms to the left:

9m^2-(3m+20)=0

We get rid of parentheses

9m^2-3m-20=0

a = 9; b = -3; c = -20;
Δ = b2-4ac
Δ = -32-4·9·(-20)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-27}{2*9}=\frac{-24}{18}
=-1+1/3
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+27}{2*9}=\frac{30}{18}
=1+2/3
$