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49x^2-4x-4=0
a = 49; b = -4; c = -4;
Δ = b2-4ac
Δ = -42-4·49·(-4)
Δ = 800
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-20\sqrt{2}}{2*49}=\frac{4-20\sqrt{2}}{98} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+20\sqrt{2}}{2*49}=\frac{4+20\sqrt{2}}{98} $
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