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2x^2+4x+4=52
We move all terms to the left:
2x^2+4x+4-(52)=0
We add all the numbers together, and all the variables
2x^2+4x-48=0
a = 2; b = 4; c = -48;
Δ = b2-4ac
Δ = 42-4·2·(-48)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-20}{2*2}=\frac{-24}{4} =-6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*2}=\frac{16}{4} =4 $
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