14x^2-90x+100=0

Solution for 14x^2-90x+100=0 equation:

14x^2-90x+100=0

a = 14; b = -90; c = +100;
Δ = b2-4ac
Δ = -902-4·14·100
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-90)-50}{2*14}=\frac{40}{28}
=1+3/7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-90)+50}{2*14}=\frac{140}{28}
=5
$