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8x^2+8=40
We move all terms to the left:
8x^2+8-(40)=0
We add all the numbers together, and all the variables
8x^2-32=0
a = 8; b = 0; c = -32;
Δ = b2-4ac
Δ = 02-4·8·(-32)
Δ = 1024
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{1024}=32$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32}{2*8}=\frac{-32}{16} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32}{2*8}=\frac{32}{16} =2 $
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