1.25x^2+34x+.8=0

Solution for 1.25x^2+34x+.8=0 equation:

1.25x^2+34x+.8=0

We add all the numbers together, and all the variables

1.25x^2+34x+0.8=0

a = 1.25; b = 34; c = +0.8;
Δ = b2-4ac
Δ = 342-4·1.25·0.8
Δ = 1152
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:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-24\sqrt{2}}{2*1.25}=\frac{-34-24\sqrt{2}}{2.5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+24\sqrt{2}}{2*1.25}=\frac{-34+24\sqrt{2}}{2.5}
$