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16x^2+137=96x
We move all terms to the left:
16x^2+137-(96x)=0
a = 16; b = -96; c = +137;
Δ = b2-4ac
Δ = -962-4·16·137
Δ = 448
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-8\sqrt{7}}{2*16}=\frac{96-8\sqrt{7}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+8\sqrt{7}}{2*16}=\frac{96+8\sqrt{7}}{32} $
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