(x-3)(2x+12)=4

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Solution for (x-3)(2x+12)=4 equation:



(x-3)(2x+12)=4
We move all terms to the left:
(x-3)(2x+12)-(4)=0
We multiply parentheses ..
(+2x^2+12x-6x-36)-4=0
We get rid of parentheses
2x^2+12x-6x-36-4=0
We add all the numbers together, and all the variables
2x^2+6x-40=0
a = 2; b = 6; c = -40;
Δ = b2-4ac
Δ = 62-4·2·(-40)
Δ = 356
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:
$\sqrt{\Delta}=\sqrt{356}=\sqrt{4*89}=\sqrt{4}*\sqrt{89}=2\sqrt{89}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{89}}{2*2}=\frac{-6-2\sqrt{89}}{4} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{89}}{2*2}=\frac{-6+2\sqrt{89}}{4} $

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