-9x^2+20x+3=0

Solution for -9x^2+20x+3=0 equation:

-9x^2+20x+3=0

a = -9; b = 20; c = +3;
Δ = b2-4ac
Δ = 202-4·(-9)·3
Δ = 508
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:

$\sqrt{\Delta}=\sqrt{508}=\sqrt{4*127}=\sqrt{4}*\sqrt{127}=2\sqrt{127}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{127}}{2*-9}=\frac{-20-2\sqrt{127}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{127}}{2*-9}=\frac{-20+2\sqrt{127}}{-18}
$