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6x-(x-4)(x-4)=104-3x^2
We move all terms to the left:
6x-(x-4)(x-4)-(104-3x^2)=0
We get rid of parentheses
3x^2+6x-(x-4)(x-4)-104=0
We multiply parentheses ..
3x^2-(+x^2-4x-4x+16)+6x-104=0
We get rid of parentheses
3x^2-x^2+4x+4x+6x-16-104=0
We add all the numbers together, and all the variables
2x^2+14x-120=0
a = 2; b = 14; c = -120;
Δ = b2-4ac
Δ = 142-4·2·(-120)
Δ = 1156
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{1156}=34$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-34}{2*2}=\frac{-48}{4} =-12 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+34}{2*2}=\frac{20}{4} =5 $
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