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3x^2+24x=15
We move all terms to the left:
3x^2+24x-(15)=0
a = 3; b = 24; c = -15;
Δ = b2-4ac
Δ = 242-4·3·(-15)
Δ = 756
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{756}=\sqrt{36*21}=\sqrt{36}*\sqrt{21}=6\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-6\sqrt{21}}{2*3}=\frac{-24-6\sqrt{21}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+6\sqrt{21}}{2*3}=\frac{-24+6\sqrt{21}}{6} $
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