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(2x+1)(x-3)+6=0
We multiply parentheses ..
(+2x^2-6x+x-3)+6=0
We get rid of parentheses
2x^2-6x+x-3+6=0
We add all the numbers together, and all the variables
2x^2-5x+3=0
a = 2; b = -5; c = +3;
Δ = b2-4ac
Δ = -52-4·2·3
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1}{2*2}=\frac{4}{4} =1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*2}=\frac{6}{4} =1+1/2 $
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