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(3x-2)(x+4)=-12x
We move all terms to the left:
(3x-2)(x+4)-(-12x)=0
We get rid of parentheses
(3x-2)(x+4)+12x=0
We multiply parentheses ..
(+3x^2+12x-2x-8)+12x=0
We get rid of parentheses
3x^2+12x-2x+12x-8=0
We add all the numbers together, and all the variables
3x^2+22x-8=0
a = 3; b = 22; c = -8;
Δ = b2-4ac
Δ = 222-4·3·(-8)
Δ = 580
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{580}=\sqrt{4*145}=\sqrt{4}*\sqrt{145}=2\sqrt{145}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{145}}{2*3}=\frac{-22-2\sqrt{145}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{145}}{2*3}=\frac{-22+2\sqrt{145}}{6} $
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