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2x^2+4x+8=78
We move all terms to the left:
2x^2+4x+8-(78)=0
We add all the numbers together, and all the variables
2x^2+4x-70=0
a = 2; b = 4; c = -70;
Δ = b2-4ac
Δ = 42-4·2·(-70)
Δ = 576
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{576}=24$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-24}{2*2}=\frac{-28}{4} =-7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+24}{2*2}=\frac{20}{4} =5 $
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