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x^2+8x-16=-28
We move all terms to the left:
x^2+8x-16-(-28)=0
We add all the numbers together, and all the variables
x^2+8x+12=0
a = 1; b = 8; c = +12;
Δ = b2-4ac
Δ = 82-4·1·12
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4}{2*1}=\frac{-12}{2} =-6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*1}=\frac{-4}{2} =-2 $
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