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6x^2+3=15=4x^2+7
We move all terms to the left:
6x^2+3-(15)=0
We add all the numbers together, and all the variables
6x^2-12=0
a = 6; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·6·(-12)
Δ = 288
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{2}}{2*6}=\frac{0-12\sqrt{2}}{12} =-\frac{12\sqrt{2}}{12} =-\sqrt{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{2}}{2*6}=\frac{0+12\sqrt{2}}{12} =\frac{12\sqrt{2}}{12} =\sqrt{2} $
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