35x^2+192x+256=0

Solution for 35x^2+192x+256=0 equation:

35x^2+192x+256=0

a = 35; b = 192; c = +256;
Δ = b2-4ac
Δ = 1922-4·35·256
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(192)-32}{2*35}=\frac{-224}{70}
=-3+1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(192)+32}{2*35}=\frac{-160}{70}
=-2+2/7
$