49x^2+147x-96=0

Solution for 49x^2+147x-96=0 equation:

49x^2+147x-96=0

a = 49; b = 147; c = -96;
Δ = b2-4ac
Δ = 1472-4·49·(-96)
Δ = 40425
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:

$\sqrt{\Delta}=\sqrt{40425}=\sqrt{1225*33}=\sqrt{1225}*\sqrt{33}=35\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(147)-35\sqrt{33}}{2*49}=\frac{-147-35\sqrt{33}}{98}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(147)+35\sqrt{33}}{2*49}=\frac{-147+35\sqrt{33}}{98}
$