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(x+1)(x+2)=11+x
We move all terms to the left:
(x+1)(x+2)-(11+x)=0
We add all the numbers together, and all the variables
(x+1)(x+2)-(x+11)=0
We get rid of parentheses
(x+1)(x+2)-x-11=0
We multiply parentheses ..
(+x^2+2x+x+2)-x-11=0
We add all the numbers together, and all the variables
(+x^2+2x+x+2)-1x-11=0
We get rid of parentheses
x^2+2x+x-1x+2-11=0
We add all the numbers together, and all the variables
x^2+2x-9=0
a = 1; b = 2; c = -9;
Δ = b2-4ac
Δ = 22-4·1·(-9)
Δ = 40
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{10}}{2*1}=\frac{-2-2\sqrt{10}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{10}}{2*1}=\frac{-2+2\sqrt{10}}{2} $
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