(m+8)(m-5)=0

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Solution for (m+8)(m-5)=0 equation:



(m+8)(m-5)=0
We multiply parentheses ..
(+m^2-5m+8m-40)=0
We get rid of parentheses
m^2-5m+8m-40=0
We add all the numbers together, and all the variables
m^2+3m-40=0
a = 1; b = 3; c = -40;
Δ = b2-4ac
Δ = 32-4·1·(-40)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-13}{2*1}=\frac{-16}{2} =-8 $
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13}{2*1}=\frac{10}{2} =5 $

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