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2x^2-2=3x^2-6x+3
We move all terms to the left:
2x^2-2-(3x^2-6x+3)=0
We get rid of parentheses
2x^2-3x^2+6x-3-2=0
We add all the numbers together, and all the variables
-1x^2+6x-5=0
a = -1; b = 6; c = -5;
Δ = b2-4ac
Δ = 62-4·(-1)·(-5)
Δ = 16
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4}{2*-1}=\frac{-10}{-2} =+5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4}{2*-1}=\frac{-2}{-2} =1 $
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