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(2x-1)(x+1)=4
We move all terms to the left:
(2x-1)(x+1)-(4)=0
We multiply parentheses ..
(+2x^2+2x-1x-1)-4=0
We get rid of parentheses
2x^2+2x-1x-1-4=0
We add all the numbers together, and all the variables
2x^2+x-5=0
a = 2; b = 1; c = -5;
Δ = b2-4ac
Δ = 12-4·2·(-5)
Δ = 41
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{41}}{2*2}=\frac{-1-\sqrt{41}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{41}}{2*2}=\frac{-1+\sqrt{41}}{4} $
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