21m^2-13+2=0

Solution for 21m^2-13+2=0 equation:

21m^2-13+2=0

We add all the numbers together, and all the variables

21m^2-11=0

a = 21; b = 0; c = -11;
Δ = b2-4ac
Δ = 02-4·21·(-11)
Δ = 924
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{924}=\sqrt{4*231}=\sqrt{4}*\sqrt{231}=2\sqrt{231}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{231}}{2*21}=\frac{0-2\sqrt{231}}{42}
=-\frac{2\sqrt{231}}{42}
=-\frac{\sqrt{231}}{21}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{231}}{2*21}=\frac{0+2\sqrt{231}}{42}
=\frac{2\sqrt{231}}{42}
=\frac{\sqrt{231}}{21}
$