4m2+11=31

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Solution for 4m2+11=31 equation:



4m^2+11=31
We move all terms to the left:
4m^2+11-(31)=0
We add all the numbers together, and all the variables
4m^2-20=0
a = 4; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·4·(-20)
Δ = 320
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:
$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{5}}{2*4}=\frac{0-8\sqrt{5}}{8} =-\frac{8\sqrt{5}}{8} =-\sqrt{5} $
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{5}}{2*4}=\frac{0+8\sqrt{5}}{8} =\frac{8\sqrt{5}}{8} =\sqrt{5} $

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