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2x^2+12x+16=x^2-8x+16
We move all terms to the left:
2x^2+12x+16-(x^2-8x+16)=0
We get rid of parentheses
2x^2-x^2+12x+8x-16+16=0
We add all the numbers together, and all the variables
x^2+20x=0
a = 1; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·1·0
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*1}=\frac{-40}{2} =-20 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*1}=\frac{0}{2} =0 $
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