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8n(6n+5)=11
We move all terms to the left:
8n(6n+5)-(11)=0
We multiply parentheses
48n^2+40n-11=0
a = 48; b = 40; c = -11;
Δ = b2-4ac
Δ = 402-4·48·(-11)
Δ = 3712
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3712}=\sqrt{64*58}=\sqrt{64}*\sqrt{58}=8\sqrt{58}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{58}}{2*48}=\frac{-40-8\sqrt{58}}{96} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{58}}{2*48}=\frac{-40+8\sqrt{58}}{96} $
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