16x^2+141=96x

Solution for 16x^2+141=96x equation:

16x^2+141=96x

We move all terms to the left:

16x^2+141-(96x)=0

a = 16; b = -96; c = +141;
Δ = b2-4ac
Δ = -962-4·16·141
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-8\sqrt{3}}{2*16}=\frac{96-8\sqrt{3}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+8\sqrt{3}}{2*16}=\frac{96+8\sqrt{3}}{32}
$