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4n^2-22+10=0
We add all the numbers together, and all the variables
4n^2-12=0
a = 4; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·4·(-12)
Δ = 192
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*4}=\frac{0-8\sqrt{3}}{8} =-\frac{8\sqrt{3}}{8} =-\sqrt{3} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*4}=\frac{0+8\sqrt{3}}{8} =\frac{8\sqrt{3}}{8} =\sqrt{3} $
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