45x^2+99x+18=0

Solution for 45x^2+99x+18=0 equation:

45x^2+99x+18=0

a = 45; b = 99; c = +18;
Δ = b2-4ac
Δ = 992-4·45·18
Δ = 6561
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6561}=81$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(99)-81}{2*45}=\frac{-180}{90}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(99)+81}{2*45}=\frac{-18}{90}
=-1/5
$