x(x+1)-8=(x+4)-5x

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Solution for x(x+1)-8=(x+4)-5x equation:



x(x+1)-8=(x+4)-5x
We move all terms to the left:
x(x+1)-8-((x+4)-5x)=0
We multiply parentheses
x^2+x-((x+4)-5x)-8=0
We calculate terms in parentheses: -((x+4)-5x), so:
(x+4)-5x
We add all the numbers together, and all the variables
-5x+(x+4)
We get rid of parentheses
-5x+x+4
We add all the numbers together, and all the variables
-4x+4
Back to the equation:
-(-4x+4)
We get rid of parentheses
x^2+x+4x-4-8=0
We add all the numbers together, and all the variables
x^2+5x-12=0
a = 1; b = 5; c = -12;
Δ = b2-4ac
Δ = 52-4·1·(-12)
Δ = 73
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{-(5)-\sqrt{73}}{2*1}=\frac{-5-\sqrt{73}}{2} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{73}}{2*1}=\frac{-5+\sqrt{73}}{2} $

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