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32n^2+30n-4608=0
a = 32; b = 30; c = -4608;
Δ = b2-4ac
Δ = 302-4·32·(-4608)
Δ = 590724
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{590724}=\sqrt{36*16409}=\sqrt{36}*\sqrt{16409}=6\sqrt{16409}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-6\sqrt{16409}}{2*32}=\frac{-30-6\sqrt{16409}}{64} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+6\sqrt{16409}}{2*32}=\frac{-30+6\sqrt{16409}}{64} $
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