4x^2+70x-500=0

Solution for 4x^2+70x-500=0 equation:

4x^2+70x-500=0

a = 4; b = 70; c = -500;
Δ = b2-4ac
Δ = 702-4·4·(-500)
Δ = 12900
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:

$\sqrt{\Delta}=\sqrt{12900}=\sqrt{100*129}=\sqrt{100}*\sqrt{129}=10\sqrt{129}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(70)-10\sqrt{129}}{2*4}=\frac{-70-10\sqrt{129}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(70)+10\sqrt{129}}{2*4}=\frac{-70+10\sqrt{129}}{8}
$