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(2x+1)(x-3)=-4
We move all terms to the left:
(2x+1)(x-3)-(-4)=0
We add all the numbers together, and all the variables
(2x+1)(x-3)+4=0
We multiply parentheses ..
(+2x^2-6x+x-3)+4=0
We get rid of parentheses
2x^2-6x+x-3+4=0
We add all the numbers together, and all the variables
2x^2-5x+1=0
a = 2; b = -5; c = +1;
Δ = b2-4ac
Δ = -52-4·2·1
Δ = 17
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{-(-5)-\sqrt{17}}{2*2}=\frac{5-\sqrt{17}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{17}}{2*2}=\frac{5+\sqrt{17}}{4} $
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