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(3x)(2x+4)=378
We move all terms to the left:
(3x)(2x+4)-(378)=0
We multiply parentheses
6x^2+12x-378=0
a = 6; b = 12; c = -378;
Δ = b2-4ac
Δ = 122-4·6·(-378)
Δ = 9216
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{9216}=96$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-96}{2*6}=\frac{-108}{12} =-9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+96}{2*6}=\frac{84}{12} =7 $
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