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4x+5=(x+1)(x+2)
We move all terms to the left:
4x+5-((x+1)(x+2))=0
We multiply parentheses ..
-((+x^2+2x+x+2))+4x+5=0
We calculate terms in parentheses: -((+x^2+2x+x+2)), so:We add all the numbers together, and all the variables
(+x^2+2x+x+2)
We get rid of parentheses
x^2+2x+x+2
We add all the numbers together, and all the variables
x^2+3x+2
Back to the equation:
-(x^2+3x+2)
4x-(x^2+3x+2)+5=0
We get rid of parentheses
-x^2+4x-3x-2+5=0
We add all the numbers together, and all the variables
-1x^2+x+3=0
a = -1; b = 1; c = +3;
Δ = b2-4ac
Δ = 12-4·(-1)·3
Δ = 13
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{13}}{2*-1}=\frac{-1-\sqrt{13}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{13}}{2*-1}=\frac{-1+\sqrt{13}}{-2} $
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