10x2+20x-13=0

Solution for 10x2+20x-13=0 equation:

10x^2+20x-13=0

a = 10; b = 20; c = -13;
Δ = b2-4ac
Δ = 202-4·10·(-13)
Δ = 920
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:

$\sqrt{\Delta}=\sqrt{920}=\sqrt{4*230}=\sqrt{4}*\sqrt{230}=2\sqrt{230}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{230}}{2*10}=\frac{-20-2\sqrt{230}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{230}}{2*10}=\frac{-20+2\sqrt{230}}{20}
$