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X+4X^2=8
We move all terms to the left:
X+4X^2-(8)=0
a = 4; b = 1; c = -8;
Δ = b2-4ac
Δ = 12-4·4·(-8)
Δ = 129
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{129}}{2*4}=\frac{-1-\sqrt{129}}{8} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{129}}{2*4}=\frac{-1+\sqrt{129}}{8} $
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