3x^2-18x-49=0

Solution for 3x^2-18x-49=0 equation:

3x^2-18x-49=0

a = 3; b = -18; c = -49;
Δ = b2-4ac
Δ = -182-4·3·(-49)
Δ = 912
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:

$\sqrt{\Delta}=\sqrt{912}=\sqrt{16*57}=\sqrt{16}*\sqrt{57}=4\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-4\sqrt{57}}{2*3}=\frac{18-4\sqrt{57}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+4\sqrt{57}}{2*3}=\frac{18+4\sqrt{57}}{6}
$