5x^2+4x-119=0

Solution for 5x^2+4x-119=0 equation:

5x^2+4x-119=0

a = 5; b = 4; c = -119;
Δ = b2-4ac
Δ = 42-4·5·(-119)
Δ = 2396
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:

$\sqrt{\Delta}=\sqrt{2396}=\sqrt{4*599}=\sqrt{4}*\sqrt{599}=2\sqrt{599}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{599}}{2*5}=\frac{-4-2\sqrt{599}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{599}}{2*5}=\frac{-4+2\sqrt{599}}{10}
$