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39x^2+4x-4=0
a = 39; b = 4; c = -4;
Δ = b2-4ac
Δ = 42-4·39·(-4)
Δ = 640
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8\sqrt{10}}{2*39}=\frac{-4-8\sqrt{10}}{78} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8\sqrt{10}}{2*39}=\frac{-4+8\sqrt{10}}{78} $
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