8m-20+m^2=0

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Solution for 8m-20+m^2=0 equation: 



8m-20+m^2=0

a = 1; b = 8; c = -20;
Δ = b2-4ac
Δ = 82-4·1·(-20)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$


$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-12}{2*1}=\frac{-20}{2}
=-10
$


$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+12}{2*1}=\frac{4}{2}
=2
$

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