7z^2+49z=0

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Solution for 7z^2+49z=0 equation: 



7z^2+49z=0

a = 7; b = 49; c = 0;
Δ = b2-4ac
Δ = 492-4·7·0
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$


$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-49}{2*7}=\frac{-98}{14}
=-7
$


$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+49}{2*7}=\frac{0}{14}
=0
$

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