(2x+1)(3x-5)+(8x+3)(2x+1)=0

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Solution for (2x+1)(3x-5)+(8x+3)(2x+1)=0 equation:



(2x+1)(3x-5)+(8x+3)(2x+1)=0
We multiply parentheses ..
(+6x^2-10x+3x-5)+(8x+3)(2x+1)=0
We get rid of parentheses
6x^2-10x+3x+(8x+3)(2x+1)-5=0
We multiply parentheses ..
6x^2+(+16x^2+8x+6x+3)-10x+3x-5=0
We add all the numbers together, and all the variables
6x^2+(+16x^2+8x+6x+3)-7x-5=0
We get rid of parentheses
6x^2+16x^2+8x+6x-7x+3-5=0
We add all the numbers together, and all the variables
22x^2+7x-2=0
a = 22; b = 7; c = -2;
Δ = b2-4ac
Δ = 72-4·22·(-2)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-15}{2*22}=\frac{-22}{44} =-1/2 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+15}{2*22}=\frac{8}{44} =2/11 $

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