-4c^2+14c=0

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Solution for -4c^2+14c=0 equation:



-4c^2+14c=0
a = -4; b = 14; c = 0;
Δ = b2-4ac
Δ = 142-4·(-4)·0
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-14}{2*-4}=\frac{-28}{-8} =3+1/2 $
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14}{2*-4}=\frac{0}{-8} =0 $

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