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16x^2-108x+120=0
a = 16; b = -108; c = +120;
Δ = b2-4ac
Δ = -1082-4·16·120
Δ = 3984
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3984}=\sqrt{16*249}=\sqrt{16}*\sqrt{249}=4\sqrt{249}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-108)-4\sqrt{249}}{2*16}=\frac{108-4\sqrt{249}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-108)+4\sqrt{249}}{2*16}=\frac{108+4\sqrt{249}}{32} $
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