(5+2x)(4+2x)-20=22

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Solution for (5+2x)(4+2x)-20=22 equation:



(5+2x)(4+2x)-20=22
We move all terms to the left:
(5+2x)(4+2x)-20-(22)=0
We add all the numbers together, and all the variables
(2x+5)(2x+4)-20-22=0
We add all the numbers together, and all the variables
(2x+5)(2x+4)-42=0
We multiply parentheses ..
(+4x^2+8x+10x+20)-42=0
We get rid of parentheses
4x^2+8x+10x+20-42=0
We add all the numbers together, and all the variables
4x^2+18x-22=0
a = 4; b = 18; c = -22;
Δ = b2-4ac
Δ = 182-4·4·(-22)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-26}{2*4}=\frac{-44}{8} =-5+1/2 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+26}{2*4}=\frac{8}{8} =1 $

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