50x^2+100x-130=0

Solution for 50x^2+100x-130=0 equation:

50x^2+100x-130=0

a = 50; b = 100; c = -130;
Δ = b2-4ac
Δ = 1002-4·50·(-130)
Δ = 36000
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:

$\sqrt{\Delta}=\sqrt{36000}=\sqrt{3600*10}=\sqrt{3600}*\sqrt{10}=60\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-60\sqrt{10}}{2*50}=\frac{-100-60\sqrt{10}}{100}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+60\sqrt{10}}{2*50}=\frac{-100+60\sqrt{10}}{100}
$