m^2-42=0

Solution for m^2-42=0 equation:

m^2-42=0

a = 1; b = 0; c = -42;
Δ = b2-4ac
Δ = 02-4·1·(-42)
Δ = 168
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:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{42}}{2*1}=\frac{0-2\sqrt{42}}{2}
=-\frac{2\sqrt{42}}{2}
=-\sqrt{42}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{42}}{2*1}=\frac{0+2\sqrt{42}}{2}
=\frac{2\sqrt{42}}{2}
=\sqrt{42}
$