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(2x+1)(x+3)=4
We move all terms to the left:
(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+7x-1=0
a = 2; b = 7; c = -1;
Δ = b2-4ac
Δ = 72-4·2·(-1)
Δ = 57
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{-(7)-\sqrt{57}}{2*2}=\frac{-7-\sqrt{57}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{57}}{2*2}=\frac{-7+\sqrt{57}}{4} $
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