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2X^2+4X=3+3X^2
We move all terms to the left:
2X^2+4X-(3+3X^2)=0
We get rid of parentheses
2X^2-3X^2+4X-3=0
We add all the numbers together, and all the variables
-1X^2+4X-3=0
a = -1; b = 4; c = -3;
Δ = b2-4ac
Δ = 42-4·(-1)·(-3)
Δ = 4
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{4}=2$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2}{2*-1}=\frac{-6}{-2} =+3 $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2}{2*-1}=\frac{-2}{-2} =1 $
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