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2x^2+4x+4=100
We move all terms to the left:
2x^2+4x+4-(100)=0
We add all the numbers together, and all the variables
2x^2+4x-96=0
a = 2; b = 4; c = -96;
Δ = b2-4ac
Δ = 42-4·2·(-96)
Δ = 784
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{784}=28$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-28}{2*2}=\frac{-32}{4} =-8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+28}{2*2}=\frac{24}{4} =6 $
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