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3x^2+18x=45
We move all terms to the left:
3x^2+18x-(45)=0
a = 3; b = 18; c = -45;
Δ = b2-4ac
Δ = 182-4·3·(-45)
Δ = 864
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{864}=\sqrt{144*6}=\sqrt{144}*\sqrt{6}=12\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-12\sqrt{6}}{2*3}=\frac{-18-12\sqrt{6}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+12\sqrt{6}}{2*3}=\frac{-18+12\sqrt{6}}{6} $
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