0.4x^2+28x-245=0

Solution for 0.4x^2+28x-245=0 equation:

0.4x^2+28x-245=0

a = 0.4; b = 28; c = -245;
Δ = b2-4ac
Δ = 282-4·0.4·(-245)
Δ = 1176
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:

$\sqrt{\Delta}=\sqrt{1176}=\sqrt{196*6}=\sqrt{196}*\sqrt{6}=14\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-14\sqrt{6}}{2*0.4}=\frac{-28-14\sqrt{6}}{0.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+14\sqrt{6}}{2*0.4}=\frac{-28+14\sqrt{6}}{0.8}
$