9x^2+42x+27=0

Solution for 9x^2+42x+27=0 equation:

9x^2+42x+27=0

a = 9; b = 42; c = +27;
Δ = b2-4ac
Δ = 422-4·9·27
Δ = 792
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:

$\sqrt{\Delta}=\sqrt{792}=\sqrt{36*22}=\sqrt{36}*\sqrt{22}=6\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-6\sqrt{22}}{2*9}=\frac{-42-6\sqrt{22}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+6\sqrt{22}}{2*9}=\frac{-42+6\sqrt{22}}{18}
$