5x^2+40x-242=0

Solution for 5x^2+40x-242=0 equation:

5x^2+40x-242=0

a = 5; b = 40; c = -242;
Δ = b2-4ac
Δ = 402-4·5·(-242)
Δ = 6440
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:

$\sqrt{\Delta}=\sqrt{6440}=\sqrt{4*1610}=\sqrt{4}*\sqrt{1610}=2\sqrt{1610}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{1610}}{2*5}=\frac{-40-2\sqrt{1610}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{1610}}{2*5}=\frac{-40+2\sqrt{1610}}{10}
$