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0=2+40x-16x^2
We move all terms to the left:
0-(2+40x-16x^2)=0
We add all the numbers together, and all the variables
-(2+40x-16x^2)=0
We get rid of parentheses
16x^2-40x-2=0
a = 16; b = -40; c = -2;
Δ = b2-4ac
Δ = -402-4·16·(-2)
Δ = 1728
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1728}=\sqrt{576*3}=\sqrt{576}*\sqrt{3}=24\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-24\sqrt{3}}{2*16}=\frac{40-24\sqrt{3}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+24\sqrt{3}}{2*16}=\frac{40+24\sqrt{3}}{32} $
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