x2-142x+96=0

Solution for x2-142x+96=0 equation:

x2-142x+96=0

We add all the numbers together, and all the variables

x^2-142x+96=0

a = 1; b = -142; c = +96;
Δ = b2-4ac
Δ = -1422-4·1·96
Δ = 19780
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:

$\sqrt{\Delta}=\sqrt{19780}=\sqrt{4*4945}=\sqrt{4}*\sqrt{4945}=2\sqrt{4945}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-142)-2\sqrt{4945}}{2*1}=\frac{142-2\sqrt{4945}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-142)+2\sqrt{4945}}{2*1}=\frac{142+2\sqrt{4945}}{2}
$