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X2+4X^2=16
We move all terms to the left:
X2+4X^2-(16)=0
We add all the numbers together, and all the variables
5X^2-16=0
a = 5; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·5·(-16)
Δ = 320
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{5}}{2*5}=\frac{0-8\sqrt{5}}{10} =-\frac{8\sqrt{5}}{10} =-\frac{4\sqrt{5}}{5} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{5}}{2*5}=\frac{0+8\sqrt{5}}{10} =\frac{8\sqrt{5}}{10} =\frac{4\sqrt{5}}{5} $
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