-2m^2+19m-24=0

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Solution for -2m^2+19m-24=0 equation:



-2m^2+19m-24=0
a = -2; b = 19; c = -24;
Δ = b2-4ac
Δ = 192-4·(-2)·(-24)
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{-(19)-13}{2*-2}=\frac{-32}{-4} =+8 $
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*-2}=\frac{-6}{-4} =1+1/2 $

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