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(11x+4)2x=840
We move all terms to the left:
(11x+4)2x-(840)=0
We multiply parentheses
22x^2+8x-840=0
a = 22; b = 8; c = -840;
Δ = b2-4ac
Δ = 82-4·22·(-840)
Δ = 73984
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{73984}=272x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-272}{2*22}=\frac{-280}{44} =-6+4/11x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+272}{2*22}=\frac{264}{44} =6
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