50x^2+39x+7=0

Solution for 50x^2+39x+7=0 equation:

50x^2+39x+7=0

a = 50; b = 39; c = +7;
Δ = b2-4ac
Δ = 392-4·50·7
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-11}{2*50}=\frac{-50}{100}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+11}{2*50}=\frac{-28}{100}
=-7/25
$