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12x^2+39x+30x=3
We move all terms to the left:
12x^2+39x+30x-(3)=0
We add all the numbers together, and all the variables
12x^2+69x-3=0
a = 12; b = 69; c = -3;
Δ = b2-4ac
Δ = 692-4·12·(-3)
Δ = 4905
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4905}=\sqrt{9*545}=\sqrt{9}*\sqrt{545}=3\sqrt{545}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(69)-3\sqrt{545}}{2*12}=\frac{-69-3\sqrt{545}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(69)+3\sqrt{545}}{2*12}=\frac{-69+3\sqrt{545}}{24} $
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