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2x^2+5x+-3=x2+4x+3
We move all terms to the left:
2x^2+5x+-3-(x2+4x+3)=0
We add all the numbers together, and all the variables
2x^2-(+x^2+4x+3)+5x-3+=0
We add all the numbers together, and all the variables
2x^2-(+x^2+4x+3)+5x=0
We get rid of parentheses
2x^2-x^2-4x+5x-3=0
We add all the numbers together, and all the variables
x^2+x-3=0
a = 1; b = 1; c = -3;
Δ = b2-4ac
Δ = 12-4·1·(-3)
Δ = 13
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{13}}{2*1}=\frac{-1-\sqrt{13}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{13}}{2*1}=\frac{-1+\sqrt{13}}{2} $
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