-28r+4r^2=0

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Solution for -28r+4r^2=0 equation: 



-28r+4r^2=0

a = 4; b = -28; c = 0;
Δ = b2-4ac
Δ = -282-4·4·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-28}{2*4}=\frac{0}{8}
=0
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28}{2*4}=\frac{56}{8}
=7
$

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