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(x+3)(2x-4)=5
We move all terms to the left:
(x+3)(2x-4)-(5)=0
We multiply parentheses ..
(+2x^2-4x+6x-12)-5=0
We get rid of parentheses
2x^2-4x+6x-12-5=0
We add all the numbers together, and all the variables
2x^2+2x-17=0
a = 2; b = 2; c = -17;
Δ = b2-4ac
Δ = 22-4·2·(-17)
Δ = 140
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{35}}{2*2}=\frac{-2-2\sqrt{35}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{35}}{2*2}=\frac{-2+2\sqrt{35}}{4} $
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