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3x+10=x^2-6x
We move all terms to the left:
3x+10-(x^2-6x)=0
We get rid of parentheses
-x^2+3x+6x+10=0
We add all the numbers together, and all the variables
-1x^2+9x+10=0
a = -1; b = 9; c = +10;
Δ = b2-4ac
Δ = 92-4·(-1)·10
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-11}{2*-1}=\frac{-20}{-2} =+10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+11}{2*-1}=\frac{2}{-2} =-1 $
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