4x^2+20x-7150=0

Solution for 4x^2+20x-7150=0 equation:

4x^2+20x-7150=0

a = 4; b = 20; c = -7150;
Δ = b2-4ac
Δ = 202-4·4·(-7150)
Δ = 114800
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:

$\sqrt{\Delta}=\sqrt{114800}=\sqrt{400*287}=\sqrt{400}*\sqrt{287}=20\sqrt{287}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20\sqrt{287}}{2*4}=\frac{-20-20\sqrt{287}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20\sqrt{287}}{2*4}=\frac{-20+20\sqrt{287}}{8}
$