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12x^2-24=4x+80
We move all terms to the left:
12x^2-24-(4x+80)=0
We get rid of parentheses
12x^2-4x-80-24=0
We add all the numbers together, and all the variables
12x^2-4x-104=0
a = 12; b = -4; c = -104;
Δ = b2-4ac
Δ = -42-4·12·(-104)
Δ = 5008
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5008}=\sqrt{16*313}=\sqrt{16}*\sqrt{313}=4\sqrt{313}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{313}}{2*12}=\frac{4-4\sqrt{313}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{313}}{2*12}=\frac{4+4\sqrt{313}}{24} $
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