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12x^2-96x+144=0
a = 12; b = -96; c = +144;
Δ = b2-4ac
Δ = -962-4·12·144
Δ = 2304
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2304}=48$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-48}{2*12}=\frac{48}{24} =2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+48}{2*12}=\frac{144}{24} =6 $
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