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5x-10=2-2x(x-3)
We move all terms to the left:
5x-10-(2-2x(x-3))=0
We calculate terms in parentheses: -(2-2x(x-3)), so:We get rid of parentheses
2-2x(x-3)
determiningTheFunctionDomain -2x(x-3)+2
We multiply parentheses
-2x^2+6x+2
Back to the equation:
-(-2x^2+6x+2)
2x^2-6x+5x-2-10=0
We add all the numbers together, and all the variables
2x^2-1x-12=0
a = 2; b = -1; c = -12;
Δ = b2-4ac
Δ = -12-4·2·(-12)
Δ = 97
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{97}}{2*2}=\frac{1-\sqrt{97}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{97}}{2*2}=\frac{1+\sqrt{97}}{4} $
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