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6(x+4)+2(x+1)=(x+1)(x+4)
We move all terms to the left:
6(x+4)+2(x+1)-((x+1)(x+4))=0
We multiply parentheses
6x+2x-((x+1)(x+4))+24+2=0
We multiply parentheses ..
-((+x^2+4x+x+4))+6x+2x+24+2=0
We calculate terms in parentheses: -((+x^2+4x+x+4)), so:We add all the numbers together, and all the variables
(+x^2+4x+x+4)
We get rid of parentheses
x^2+4x+x+4
We add all the numbers together, and all the variables
x^2+5x+4
Back to the equation:
-(x^2+5x+4)
8x-(x^2+5x+4)+26=0
We get rid of parentheses
-x^2+8x-5x-4+26=0
We add all the numbers together, and all the variables
-1x^2+3x+22=0
a = -1; b = 3; c = +22;
Δ = b2-4ac
Δ = 32-4·(-1)·22
Δ = 97
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{-(3)-\sqrt{97}}{2*-1}=\frac{-3-\sqrt{97}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{97}}{2*-1}=\frac{-3+\sqrt{97}}{-2} $
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