35x^2+195x+100=0

Solution for 35x^2+195x+100=0 equation:

35x^2+195x+100=0

a = 35; b = 195; c = +100;
Δ = b2-4ac
Δ = 1952-4·35·100
Δ = 24025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{24025}=155$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(195)-155}{2*35}=\frac{-350}{70}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(195)+155}{2*35}=\frac{-40}{70}
=-4/7
$