3x^2+6x-57=0

Solution for 3x^2+6x-57=0 equation:

3x^2+6x-57=0

a = 3; b = 6; c = -57;
Δ = b2-4ac
Δ = 62-4·3·(-57)
Δ = 720
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:

$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-12\sqrt{5}}{2*3}=\frac{-6-12\sqrt{5}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+12\sqrt{5}}{2*3}=\frac{-6+12\sqrt{5}}{6}
$