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3x-10=x^2-6x+10
We move all terms to the left:
3x-10-(x^2-6x+10)=0
We get rid of parentheses
-x^2+3x+6x-10-10=0
We add all the numbers together, and all the variables
-1x^2+9x-20=0
a = -1; b = 9; c = -20;
Δ = b2-4ac
Δ = 92-4·(-1)·(-20)
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-1}{2*-1}=\frac{-10}{-2} =+5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+1}{2*-1}=\frac{-8}{-2} =+4 $
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