24x^2+13x=41

Solution for 24x^2+13x=41 equation:

24x^2+13x=41

We move all terms to the left:

24x^2+13x-(41)=0

a = 24; b = 13; c = -41;
Δ = b2-4ac
Δ = 132-4·24·(-41)
Δ = 4105
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-\sqrt{4105}}{2*24}=\frac{-13-\sqrt{4105}}{48}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{4105}}{2*24}=\frac{-13+\sqrt{4105}}{48}
$