10x^2+20x-9=0

Solution for 10x^2+20x-9=0 equation:

10x^2+20x-9=0

a = 10; b = 20; c = -9;
Δ = b2-4ac
Δ = 202-4·10·(-9)
Δ = 760
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:

$\sqrt{\Delta}=\sqrt{760}=\sqrt{4*190}=\sqrt{4}*\sqrt{190}=2\sqrt{190}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{190}}{2*10}=\frac{-20-2\sqrt{190}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{190}}{2*10}=\frac{-20+2\sqrt{190}}{20}
$