4m^2+23m+32=0

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Solution for 4m^2+23m+32=0 equation: 



4m^2+23m+32=0

a = 4; b = 23; c = +32;
Δ = b2-4ac
Δ = 232-4·4·32
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-\sqrt{17}}{2*4}=\frac{-23-\sqrt{17}}{8}
$


$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{17}}{2*4}=\frac{-23+\sqrt{17}}{8}
$

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