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x(2x+8)=42
We move all terms to the left:
x(2x+8)-(42)=0
We multiply parentheses
2x^2+8x-42=0
a = 2; b = 8; c = -42;
Δ = b2-4ac
Δ = 82-4·2·(-42)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-20}{2*2}=\frac{-28}{4} =-7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+20}{2*2}=\frac{12}{4} =3 $
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