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12x^2+28x=120
We move all terms to the left:
12x^2+28x-(120)=0
a = 12; b = 28; c = -120;
Δ = b2-4ac
Δ = 282-4·12·(-120)
Δ = 6544
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{6544}=\sqrt{16*409}=\sqrt{16}*\sqrt{409}=4\sqrt{409}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{409}}{2*12}=\frac{-28-4\sqrt{409}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{409}}{2*12}=\frac{-28+4\sqrt{409}}{24} $
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