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3x^2+2x=24
We move all terms to the left:
3x^2+2x-(24)=0
a = 3; b = 2; c = -24;
Δ = b2-4ac
Δ = 22-4·3·(-24)
Δ = 292
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{73}}{2*3}=\frac{-2-2\sqrt{73}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{73}}{2*3}=\frac{-2+2\sqrt{73}}{6} $
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