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X^2+8X+40=40
We move all terms to the left:
X^2+8X+40-(40)=0
We add all the numbers together, and all the variables
X^2+8X=0
a = 1; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·1·0
Δ = 64
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{64}=8$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*1}=\frac{-16}{2} =-8 $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*1}=\frac{0}{2} =0 $
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