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3x^2-12=2x^2+3
We move all terms to the left:
3x^2-12-(2x^2+3)=0
We get rid of parentheses
3x^2-2x^2-3-12=0
We add all the numbers together, and all the variables
x^2-15=0
a = 1; b = 0; c = -15;
Δ = b2-4ac
Δ = 02-4·1·(-15)
Δ = 60
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{15}}{2*1}=\frac{0-2\sqrt{15}}{2} =-\frac{2\sqrt{15}}{2} =-\sqrt{15} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{15}}{2*1}=\frac{0+2\sqrt{15}}{2} =\frac{2\sqrt{15}}{2} =\sqrt{15} $
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