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8x^2+40x=-2
We move all terms to the left:
8x^2+40x-(-2)=0
We add all the numbers together, and all the variables
8x^2+40x+2=0
a = 8; b = 40; c = +2;
Δ = b2-4ac
Δ = 402-4·8·2
Δ = 1536
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-16\sqrt{6}}{2*8}=\frac{-40-16\sqrt{6}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+16\sqrt{6}}{2*8}=\frac{-40+16\sqrt{6}}{16} $
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