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12x^2+48x-192=0
a = 12; b = 48; c = -192;
Δ = b2-4ac
Δ = 482-4·12·(-192)
Δ = 11520
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{11520}=\sqrt{2304*5}=\sqrt{2304}*\sqrt{5}=48\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-48\sqrt{5}}{2*12}=\frac{-48-48\sqrt{5}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48\sqrt{5}}{2*12}=\frac{-48+48\sqrt{5}}{24} $
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