5x^2+27x-432=0

Solution for 5x^2+27x-432=0 equation:

5x^2+27x-432=0

a = 5; b = 27; c = -432;
Δ = b2-4ac
Δ = 272-4·5·(-432)
Δ = 9369
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:

$\sqrt{\Delta}=\sqrt{9369}=\sqrt{9*1041}=\sqrt{9}*\sqrt{1041}=3\sqrt{1041}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-3\sqrt{1041}}{2*5}=\frac{-27-3\sqrt{1041}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+3\sqrt{1041}}{2*5}=\frac{-27+3\sqrt{1041}}{10}
$