m^2+4m-87=9

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Solution for m^2+4m-87=9 equation:



m^2+4m-87=9
We move all terms to the left:
m^2+4m-87-(9)=0
We add all the numbers together, and all the variables
m^2+4m-96=0
a = 1; b = 4; c = -96;
Δ = b2-4ac
Δ = 42-4·1·(-96)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-20}{2*1}=\frac{-24}{2} =-12 $
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*1}=\frac{16}{2} =8 $

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