-4m2+100=0

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Solution for -4m2+100=0 equation:



-4m^2+100=0
a = -4; b = 0; c = +100;
Δ = b2-4ac
Δ = 02-4·(-4)·100
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*-4}=\frac{-40}{-8} =+5 $
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*-4}=\frac{40}{-8} =-5 $

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