49x^2+28x-23=0

Solution for 49x^2+28x-23=0 equation:

49x^2+28x-23=0

a = 49; b = 28; c = -23;
Δ = b2-4ac
Δ = 282-4·49·(-23)
Δ = 5292
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:

$\sqrt{\Delta}=\sqrt{5292}=\sqrt{1764*3}=\sqrt{1764}*\sqrt{3}=42\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-42\sqrt{3}}{2*49}=\frac{-28-42\sqrt{3}}{98}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+42\sqrt{3}}{2*49}=\frac{-28+42\sqrt{3}}{98}
$