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16x^2-240x+23=0
a = 16; b = -240; c = +23;
Δ = b2-4ac
Δ = -2402-4·16·23
Δ = 56128
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{56128}=\sqrt{64*877}=\sqrt{64}*\sqrt{877}=8\sqrt{877}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-240)-8\sqrt{877}}{2*16}=\frac{240-8\sqrt{877}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-240)+8\sqrt{877}}{2*16}=\frac{240+8\sqrt{877}}{32} $
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