(4+x)(5+x)-20=10

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Solution for (4+x)(5+x)-20=10 equation:



(4+x)(5+x)-20=10
We move all terms to the left:
(4+x)(5+x)-20-(10)=0
We add all the numbers together, and all the variables
(x+4)(x+5)-20-10=0
We add all the numbers together, and all the variables
(x+4)(x+5)-30=0
We multiply parentheses ..
(+x^2+5x+4x+20)-30=0
We get rid of parentheses
x^2+5x+4x+20-30=0
We add all the numbers together, and all the variables
x^2+9x-10=0
a = 1; b = 9; c = -10;
Δ = b2-4ac
Δ = 92-4·1·(-10)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-11}{2*1}=\frac{-20}{2} =-10 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+11}{2*1}=\frac{2}{2} =1 $

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