20m^2+23m+6=0

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Solution for 20m^2+23m+6=0 equation: 



20m^2+23m+6=0

a = 20; b = 23; c = +6;
Δ = b2-4ac
Δ = 232-4·20·6
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$


$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-7}{2*20}=\frac{-30}{40}
=-3/4
$


$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+7}{2*20}=\frac{-16}{40}
=-2/5
$

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