2r^2-2r-4=0

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



2r^2-2r-4=0

a = 2; b = -2; c = -4;
Δ = b2-4ac
Δ = -22-4·2·(-4)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6}{2*2}=\frac{-4}{4}
=-1
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6}{2*2}=\frac{8}{4}
=2
$

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