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3s^2-4s-4=0
a = 3; b = -4; c = -4;
Δ = b2-4ac
Δ = -42-4·3·(-4)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{64}=8$$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*3}=\frac{-4}{6} =-2/3 $$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*3}=\frac{12}{6} =2 $
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