5x^2+20x-2464=0

Solution for 5x^2+20x-2464=0 equation:

5x^2+20x-2464=0

a = 5; b = 20; c = -2464;
Δ = b2-4ac
Δ = 202-4·5·(-2464)
Δ = 49680
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{49680}=\sqrt{144*345}=\sqrt{144}*\sqrt{345}=12\sqrt{345}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-12\sqrt{345}}{2*5}=\frac{-20-12\sqrt{345}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+12\sqrt{345}}{2*5}=\frac{-20+12\sqrt{345}}{10}
$