3x^2+56x+80=0

Solution for 3x^2+56x+80=0 equation:

3x^2+56x+80=0

a = 3; b = 56; c = +80;
Δ = b2-4ac
Δ = 562-4·3·80
Δ = 2176
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:

$\sqrt{\Delta}=\sqrt{2176}=\sqrt{64*34}=\sqrt{64}*\sqrt{34}=8\sqrt{34}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-8\sqrt{34}}{2*3}=\frac{-56-8\sqrt{34}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+8\sqrt{34}}{2*3}=\frac{-56+8\sqrt{34}}{6}
$