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4n^2+4n-10=0
a = 4; b = 4; c = -10;
Δ = b2-4ac
Δ = 42-4·4·(-10)
Δ = 176
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{11}}{2*4}=\frac{-4-4\sqrt{11}}{8} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{11}}{2*4}=\frac{-4+4\sqrt{11}}{8} $
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