4x(2x+3)=16x+3-8x+9

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Solution for 4x(2x+3)=16x+3-8x+9 equation:



4x(2x+3)=16x+3-8x+9
We move all terms to the left:
4x(2x+3)-(16x+3-8x+9)=0
We add all the numbers together, and all the variables
4x(2x+3)-(8x+12)=0
We multiply parentheses
8x^2+12x-(8x+12)=0
We get rid of parentheses
8x^2+12x-8x-12=0
We add all the numbers together, and all the variables
8x^2+4x-12=0
a = 8; b = 4; c = -12;
Δ = b2-4ac
Δ = 42-4·8·(-12)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-20}{2*8}=\frac{-24}{16} =-1+1/2 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*8}=\frac{16}{16} =1 $

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