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(x-3)(x+2)=x-3
We move all terms to the left:
(x-3)(x+2)-(x-3)=0
We get rid of parentheses
(x-3)(x+2)-x+3=0
We multiply parentheses ..
(+x^2+2x-3x-6)-x+3=0
We add all the numbers together, and all the variables
(+x^2+2x-3x-6)-1x+3=0
We get rid of parentheses
x^2+2x-3x-1x-6+3=0
We add all the numbers together, and all the variables
x^2-2x-3=0
a = 1; b = -2; c = -3;
Δ = b2-4ac
Δ = -22-4·1·(-3)
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4}{2*1}=\frac{-2}{2} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4}{2*1}=\frac{6}{2} =3 $
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