3x^2+32x-243=0

Solution for 3x^2+32x-243=0 equation:

3x^2+32x-243=0

a = 3; b = 32; c = -243;
Δ = b2-4ac
Δ = 322-4·3·(-243)
Δ = 3940
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:

$\sqrt{\Delta}=\sqrt{3940}=\sqrt{4*985}=\sqrt{4}*\sqrt{985}=2\sqrt{985}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-2\sqrt{985}}{2*3}=\frac{-32-2\sqrt{985}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+2\sqrt{985}}{2*3}=\frac{-32+2\sqrt{985}}{6}
$