16x^2-102x+140=0

Solution for 16x^2-102x+140=0 equation:

16x^2-102x+140=0

a = 16; b = -102; c = +140;
Δ = b2-4ac
Δ = -1022-4·16·140
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1444}=38$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-102)-38}{2*16}=\frac{64}{32}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-102)+38}{2*16}=\frac{140}{32}
=4+3/8
$