10m^2=41+45

Solution for 10m^2=41+45 equation:

10m^2=41+45

We move all terms to the left:

10m^2-(41+45)=0

We add all the numbers together, and all the variables

10m^2-86=0

a = 10; b = 0; c = -86;
Δ = b2-4ac
Δ = 02-4·10·(-86)
Δ = 3440
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:

$\sqrt{\Delta}=\sqrt{3440}=\sqrt{16*215}=\sqrt{16}*\sqrt{215}=4\sqrt{215}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{215}}{2*10}=\frac{0-4\sqrt{215}}{20}
=-\frac{4\sqrt{215}}{20}
=-\frac{\sqrt{215}}{5}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{215}}{2*10}=\frac{0+4\sqrt{215}}{20}
=\frac{4\sqrt{215}}{20}
=\frac{\sqrt{215}}{5}
$