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(2x-3)(3x+1)=-8x-1
We move all terms to the left:
(2x-3)(3x+1)-(-8x-1)=0
We get rid of parentheses
(2x-3)(3x+1)+8x+1=0
We multiply parentheses ..
(+6x^2+2x-9x-3)+8x+1=0
We get rid of parentheses
6x^2+2x-9x+8x-3+1=0
We add all the numbers together, and all the variables
6x^2+x-2=0
a = 6; b = 1; c = -2;
Δ = b2-4ac
Δ = 12-4·6·(-2)
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*6}=\frac{-8}{12} =-2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*6}=\frac{6}{12} =1/2 $
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