2m^2+17m=0

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Solution for 2m^2+17m=0 equation: 



2m^2+17m=0

a = 2; b = 17; c = 0;
Δ = b2-4ac
Δ = 172-4·2·0
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$


$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-17}{2*2}=\frac{-34}{4}
=-8+1/2
$


$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+17}{2*2}=\frac{0}{4}
=0
$

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