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4n^2+6n=105
We move all terms to the left:
4n^2+6n-(105)=0
a = 4; b = 6; c = -105;
Δ = b2-4ac
Δ = 62-4·4·(-105)
Δ = 1716
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1716}=\sqrt{4*429}=\sqrt{4}*\sqrt{429}=2\sqrt{429}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{429}}{2*4}=\frac{-6-2\sqrt{429}}{8} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{429}}{2*4}=\frac{-6+2\sqrt{429}}{8} $
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