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(2x+1)(3x-2)+5x(2x+1)=0
We multiply parentheses
10x^2+(2x+1)(3x-2)+5x=0
We multiply parentheses ..
10x^2+(+6x^2-4x+3x-2)+5x=0
We get rid of parentheses
10x^2+6x^2-4x+3x+5x-2=0
We add all the numbers together, and all the variables
16x^2+4x-2=0
a = 16; b = 4; c = -2;
Δ = b2-4ac
Δ = 42-4·16·(-2)
Δ = 144
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{144}=12$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12}{2*16}=\frac{-16}{32} =-1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*16}=\frac{8}{32} =1/4 $
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