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36=(x-2)(x-2+x+4)
We move all terms to the left:
36-((x-2)(x-2+x+4))=0
We add all the numbers together, and all the variables
-((x-2)(2x+2))+36=0
We multiply parentheses ..
-((+2x^2+2x-4x-4))+36=0
We calculate terms in parentheses: -((+2x^2+2x-4x-4)), so:We get rid of parentheses
(+2x^2+2x-4x-4)
We get rid of parentheses
2x^2+2x-4x-4
We add all the numbers together, and all the variables
2x^2-2x-4
Back to the equation:
-(2x^2-2x-4)
-2x^2+2x+4+36=0
We add all the numbers together, and all the variables
-2x^2+2x+40=0
a = -2; b = 2; c = +40;
Δ = b2-4ac
Δ = 22-4·(-2)·40
Δ = 324
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{324}=18$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-18}{2*-2}=\frac{-20}{-4} =+5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+18}{2*-2}=\frac{16}{-4} =-4 $
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