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2x(3x+6)=42
We move all terms to the left:
2x(3x+6)-(42)=0
We multiply parentheses
6x^2+12x-42=0
a = 6; b = 12; c = -42;
Δ = b2-4ac
Δ = 122-4·6·(-42)
Δ = 1152
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-24\sqrt{2}}{2*6}=\frac{-12-24\sqrt{2}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+24\sqrt{2}}{2*6}=\frac{-12+24\sqrt{2}}{12} $
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