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9x^2-3=3x^2+15
We move all terms to the left:
9x^2-3-(3x^2+15)=0
We get rid of parentheses
9x^2-3x^2-15-3=0
We add all the numbers together, and all the variables
6x^2-18=0
a = 6; b = 0; c = -18;
Δ = b2-4ac
Δ = 02-4·6·(-18)
Δ = 432
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{3}}{2*6}=\frac{0-12\sqrt{3}}{12} =-\frac{12\sqrt{3}}{12} =-\sqrt{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{3}}{2*6}=\frac{0+12\sqrt{3}}{12} =\frac{12\sqrt{3}}{12} =\sqrt{3} $
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