2f^2+7f=0

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Solution for 2f^2+7f=0 equation: 



2f^2+7f=0

a = 2; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·2·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*2}=\frac{-14}{4}
=-3+1/2
$


$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*2}=\frac{0}{4}
=0
$

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