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(2x+3)(x+1)=(x+3)(x+3)
We move all terms to the left:
(2x+3)(x+1)-((x+3)(x+3))=0
We multiply parentheses ..
(+2x^2+2x+3x+3)-((x+3)(x+3))=0
We calculate terms in parentheses: -((x+3)(x+3)), so:We get rid of parentheses
(x+3)(x+3)
We multiply parentheses ..
(+x^2+3x+3x+9)
We get rid of parentheses
x^2+3x+3x+9
We add all the numbers together, and all the variables
x^2+6x+9
Back to the equation:
-(x^2+6x+9)
2x^2-x^2+2x+3x-6x+3-9=0
We add all the numbers together, and all the variables
x^2-1x-6=0
a = 1; b = -1; c = -6;
Δ = b2-4ac
Δ = -12-4·1·(-6)
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*1}=\frac{-4}{2} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*1}=\frac{6}{2} =3 $
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