2(x^2+2)-4=32

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Solution for 2(x^2+2)-4=32 equation:



2(x^2+2)-4=32
We move all terms to the left:
2(x^2+2)-4-(32)=0
We add all the numbers together, and all the variables
2(x^2+2)-36=0
We multiply parentheses
2x^2+4-36=0
We add all the numbers together, and all the variables
2x^2-32=0
a = 2; b = 0; c = -32;
Δ = b2-4ac
Δ = 02-4·2·(-32)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*2}=\frac{-16}{4} =-4 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*2}=\frac{16}{4} =4 $

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