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42x^2+8x-2=0
a = 42; b = 8; c = -2;
Δ = b2-4ac
Δ = 82-4·42·(-2)
Δ = 400
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-20}{2*42}=\frac{-28}{84} =-1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+20}{2*42}=\frac{12}{84} =1/7 $
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