3x^2-28x+40=0

Solution for 3x^2-28x+40=0 equation:

3x^2-28x+40=0

a = 3; b = -28; c = +40;
Δ = b2-4ac
Δ = -282-4·3·40
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{19}}{2*3}=\frac{28-4\sqrt{19}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{19}}{2*3}=\frac{28+4\sqrt{19}}{6}
$