8x(4x-3)+8=x+5(x+1)

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Solution for 8x(4x-3)+8=x+5(x+1) equation:



8x(4x-3)+8=x+5(x+1)
We move all terms to the left:
8x(4x-3)+8-(x+5(x+1))=0
We multiply parentheses
32x^2-24x-(x+5(x+1))+8=0
We calculate terms in parentheses: -(x+5(x+1)), so:
x+5(x+1)
We multiply parentheses
x+5x+5
We add all the numbers together, and all the variables
6x+5
Back to the equation:
-(6x+5)
We get rid of parentheses
32x^2-24x-6x-5+8=0
We add all the numbers together, and all the variables
32x^2-30x+3=0
a = 32; b = -30; c = +3;
Δ = b2-4ac
Δ = -302-4·32·3
Δ = 516
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{516}=\sqrt{4*129}=\sqrt{4}*\sqrt{129}=2\sqrt{129}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-2\sqrt{129}}{2*32}=\frac{30-2\sqrt{129}}{64} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+2\sqrt{129}}{2*32}=\frac{30+2\sqrt{129}}{64} $

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