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(x+9)(x-1)=(2x+1)(x-1)
We move all terms to the left:
(x+9)(x-1)-((2x+1)(x-1))=0
We multiply parentheses ..
(+x^2-1x+9x-9)-((2x+1)(x-1))=0
We calculate terms in parentheses: -((2x+1)(x-1)), so:We get rid of parentheses
(2x+1)(x-1)
We multiply parentheses ..
(+2x^2-2x+x-1)
We get rid of parentheses
2x^2-2x+x-1
We add all the numbers together, and all the variables
2x^2-1x-1
Back to the equation:
-(2x^2-1x-1)
x^2-2x^2-1x+9x+1x-9+1=0
We add all the numbers together, and all the variables
-1x^2+9x-8=0
a = -1; b = 9; c = -8;
Δ = b2-4ac
Δ = 92-4·(-1)·(-8)
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-7}{2*-1}=\frac{-16}{-2} =+8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+7}{2*-1}=\frac{-2}{-2} =1 $
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