121x^2+154x+40=0

Solution for 121x^2+154x+40=0 equation:

121x^2+154x+40=0

a = 121; b = 154; c = +40;
Δ = b2-4ac
Δ = 1542-4·121·40
Δ = 4356
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4356}=66$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(154)-66}{2*121}=\frac{-220}{242}
=-10/11
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(154)+66}{2*121}=\frac{-88}{242}
=-4/11
$