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3x(3x-12)=24
We move all terms to the left:
3x(3x-12)-(24)=0
We multiply parentheses
9x^2-36x-24=0
a = 9; b = -36; c = -24;
Δ = b2-4ac
Δ = -362-4·9·(-24)
Δ = 2160
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2160}=\sqrt{144*15}=\sqrt{144}*\sqrt{15}=12\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-12\sqrt{15}}{2*9}=\frac{36-12\sqrt{15}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+12\sqrt{15}}{2*9}=\frac{36+12\sqrt{15}}{18} $
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