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2x(2x+4)=12
We move all terms to the left:
2x(2x+4)-(12)=0
We multiply parentheses
4x^2+8x-12=0
a = 4; b = 8; c = -12;
Δ = b2-4ac
Δ = 82-4·4·(-12)
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-16}{2*4}=\frac{-24}{8} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16}{2*4}=\frac{8}{8} =1 $
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