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12+3x+5x^2=14+4x-5x^2
We move all terms to the left:
12+3x+5x^2-(14+4x-5x^2)=0
We get rid of parentheses
5x^2+5x^2-4x+3x-14+12=0
We add all the numbers together, and all the variables
10x^2-1x-2=0
a = 10; b = -1; c = -2;
Δ = b2-4ac
Δ = -12-4·10·(-2)
Δ = 81
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{81}=9$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-9}{2*10}=\frac{-8}{20} =-2/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+9}{2*10}=\frac{10}{20} =1/2 $
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