52x^2+97x+40=0

Solution for 52x^2+97x+40=0 equation:

52x^2+97x+40=0

a = 52; b = 97; c = +40;
Δ = b2-4ac
Δ = 972-4·52·40
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(97)-33}{2*52}=\frac{-130}{104}
=-1+1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(97)+33}{2*52}=\frac{-64}{104}
=-8/13
$