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x(2x+12)=-16
We move all terms to the left:
x(2x+12)-(-16)=0
We add all the numbers together, and all the variables
x(2x+12)+16=0
We multiply parentheses
2x^2+12x+16=0
a = 2; b = 12; c = +16;
Δ = b2-4ac
Δ = 122-4·2·16
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4}{2*2}=\frac{-16}{4} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4}{2*2}=\frac{-8}{4} =-2 $
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