3x^2+16x-27=0

Solution for 3x^2+16x-27=0 equation:

3x^2+16x-27=0

a = 3; b = 16; c = -27;
Δ = b2-4ac
Δ = 162-4·3·(-27)
Δ = 580
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:

$\sqrt{\Delta}=\sqrt{580}=\sqrt{4*145}=\sqrt{4}*\sqrt{145}=2\sqrt{145}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{145}}{2*3}=\frac{-16-2\sqrt{145}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{145}}{2*3}=\frac{-16+2\sqrt{145}}{6}
$