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

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



(3x+1)(2x-5)+(3x-1)(x+2)=0
We multiply parentheses ..
(+6x^2-15x+2x-5)+(3x-1)(x+2)=0
We get rid of parentheses
6x^2-15x+2x+(3x-1)(x+2)-5=0
We multiply parentheses ..
6x^2+(+3x^2+6x-1x-2)-15x+2x-5=0
We add all the numbers together, and all the variables
6x^2+(+3x^2+6x-1x-2)-13x-5=0
We get rid of parentheses
6x^2+3x^2+6x-1x-13x-2-5=0
We add all the numbers together, and all the variables
9x^2-8x-7=0
a = 9; b = -8; c = -7;
Δ = b2-4ac
Δ = -82-4·9·(-7)
Δ = 316
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:
$\sqrt{\Delta}=\sqrt{316}=\sqrt{4*79}=\sqrt{4}*\sqrt{79}=2\sqrt{79}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{79}}{2*9}=\frac{8-2\sqrt{79}}{18} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{79}}{2*9}=\frac{8+2\sqrt{79}}{18} $

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