45x^2+9x-5=0

Solution for 45x^2+9x-5=0 equation:

45x^2+9x-5=0

a = 45; b = 9; c = -5;
Δ = b2-4ac
Δ = 92-4·45·(-5)
Δ = 981
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:

$\sqrt{\Delta}=\sqrt{981}=\sqrt{9*109}=\sqrt{9}*\sqrt{109}=3\sqrt{109}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{109}}{2*45}=\frac{-9-3\sqrt{109}}{90}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{109}}{2*45}=\frac{-9+3\sqrt{109}}{90}
$