4m^2+10=38

Solution for 4m^2+10=38 equation:

4m^2+10=38

We move all terms to the left:

4m^2+10-(38)=0

We add all the numbers together, and all the variables

4m^2-28=0

a = 4; b = 0; c = -28;
Δ = b2-4ac
Δ = 02-4·4·(-28)
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{7}}{2*4}=\frac{0-8\sqrt{7}}{8}
=-\frac{8\sqrt{7}}{8}
=-\sqrt{7}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{7}}{2*4}=\frac{0+8\sqrt{7}}{8}
=\frac{8\sqrt{7}}{8}
=\sqrt{7}
$