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X^+2X^2-16=0
We add all the numbers together, and all the variables
2X^2+X-16=0
a = 2; b = 1; c = -16;
Δ = b2-4ac
Δ = 12-4·2·(-16)
Δ = 129
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{129}}{2*2}=\frac{-1-\sqrt{129}}{4} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{129}}{2*2}=\frac{-1+\sqrt{129}}{4} $
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