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8n(2n+7)=11
We move all terms to the left:
8n(2n+7)-(11)=0
We multiply parentheses
16n^2+56n-11=0
a = 16; b = 56; c = -11;
Δ = b2-4ac
Δ = 562-4·16·(-11)
Δ = 3840
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3840}=\sqrt{256*15}=\sqrt{256}*\sqrt{15}=16\sqrt{15}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-16\sqrt{15}}{2*16}=\frac{-56-16\sqrt{15}}{32} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+16\sqrt{15}}{2*16}=\frac{-56+16\sqrt{15}}{32} $
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