49x^2+35x=0

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Solution for 49x^2+35x=0 equation:



49x^2+35x=0
a = 49; b = 35; c = 0;
Δ = b2-4ac
Δ = 352-4·49·0
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-35}{2*49}=\frac{-70}{98} =-5/7 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+35}{2*49}=\frac{0}{98} =0 $

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