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40=(x+2)(x-1)
We move all terms to the left:
40-((x+2)(x-1))=0
We multiply parentheses ..
-((+x^2-1x+2x-2))+40=0
We calculate terms in parentheses: -((+x^2-1x+2x-2)), so:We get rid of parentheses
(+x^2-1x+2x-2)
We get rid of parentheses
x^2-1x+2x-2
We add all the numbers together, and all the variables
x^2+x-2
Back to the equation:
-(x^2+x-2)
-x^2-x+2+40=0
We add all the numbers together, and all the variables
-1x^2-1x+42=0
a = -1; b = -1; c = +42;
Δ = b2-4ac
Δ = -12-4·(-1)·42
Δ = 169
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{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-13}{2*-1}=\frac{-12}{-2} =+6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+13}{2*-1}=\frac{14}{-2} =-7 $
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