10x^2+40x-350=0

Solution for 10x^2+40x-350=0 equation:

10x^2+40x-350=0

a = 10; b = 40; c = -350;
Δ = b2-4ac
Δ = 402-4·10·(-350)
Δ = 15600
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:

$\sqrt{\Delta}=\sqrt{15600}=\sqrt{400*39}=\sqrt{400}*\sqrt{39}=20\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20\sqrt{39}}{2*10}=\frac{-40-20\sqrt{39}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20\sqrt{39}}{2*10}=\frac{-40+20\sqrt{39}}{20}
$