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3x(2x+3)=96
We move all terms to the left:
3x(2x+3)-(96)=0
We multiply parentheses
6x^2+9x-96=0
a = 6; b = 9; c = -96;
Δ = b2-4ac
Δ = 92-4·6·(-96)
Δ = 2385
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2385}=\sqrt{9*265}=\sqrt{9}*\sqrt{265}=3\sqrt{265}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{265}}{2*6}=\frac{-9-3\sqrt{265}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{265}}{2*6}=\frac{-9+3\sqrt{265}}{12} $
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