3x^2-56x+98=0

Solution for 3x^2-56x+98=0 equation:

3x^2-56x+98=0

a = 3; b = -56; c = +98;
Δ = b2-4ac
Δ = -562-4·3·98
Δ = 1960
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:

$\sqrt{\Delta}=\sqrt{1960}=\sqrt{196*10}=\sqrt{196}*\sqrt{10}=14\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-56)-14\sqrt{10}}{2*3}=\frac{56-14\sqrt{10}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-56)+14\sqrt{10}}{2*3}=\frac{56+14\sqrt{10}}{6}
$