x-15x^2+28=0

Solution for x-15x^2+28=0 equation:

x-15x^2+28=0

a = -15; b = 1; c = +28;
Δ = b2-4ac
Δ = 12-4·(-15)·28
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-41}{2*-15}=\frac{-42}{-30}
=1+2/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+41}{2*-15}=\frac{40}{-30}
=-1+1/3
$