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3x^2+24x+15=6
We move all terms to the left:
3x^2+24x+15-(6)=0
We add all the numbers together, and all the variables
3x^2+24x+9=0
a = 3; b = 24; c = +9;
Δ = b2-4ac
Δ = 242-4·3·9
Δ = 468
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-6\sqrt{13}}{2*3}=\frac{-24-6\sqrt{13}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+6\sqrt{13}}{2*3}=\frac{-24+6\sqrt{13}}{6} $
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