6x^2+8x-49=0

Solution for 6x^2+8x-49=0 equation:

6x^2+8x-49=0

a = 6; b = 8; c = -49;
Δ = b2-4ac
Δ = 82-4·6·(-49)
Δ = 1240
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:

$\sqrt{\Delta}=\sqrt{1240}=\sqrt{4*310}=\sqrt{4}*\sqrt{310}=2\sqrt{310}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{310}}{2*6}=\frac{-8-2\sqrt{310}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{310}}{2*6}=\frac{-8+2\sqrt{310}}{12}
$