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(x+2)=(x+4)(x-1)
We move all terms to the left:
(x+2)-((x+4)(x-1))=0
We get rid of parentheses
x-((x+4)(x-1))+2=0
We multiply parentheses ..
-((+x^2-1x+4x-4))+x+2=0
We calculate terms in parentheses: -((+x^2-1x+4x-4)), so:We add all the numbers together, and all the variables
(+x^2-1x+4x-4)
We get rid of parentheses
x^2-1x+4x-4
We add all the numbers together, and all the variables
x^2+3x-4
Back to the equation:
-(x^2+3x-4)
x-(x^2+3x-4)+2=0
We get rid of parentheses
-x^2+x-3x+4+2=0
We add all the numbers together, and all the variables
-1x^2-2x+6=0
a = -1; b = -2; c = +6;
Δ = b2-4ac
Δ = -22-4·(-1)·6
Δ = 28
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{7}}{2*-1}=\frac{2-2\sqrt{7}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{7}}{2*-1}=\frac{2+2\sqrt{7}}{-2} $
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