9x^2+42x-49=0

Solution for 9x^2+42x-49=0 equation:

9x^2+42x-49=0

a = 9; b = 42; c = -49;
Δ = b2-4ac
Δ = 422-4·9·(-49)
Δ = 3528
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:

$\sqrt{\Delta}=\sqrt{3528}=\sqrt{1764*2}=\sqrt{1764}*\sqrt{2}=42\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-42\sqrt{2}}{2*9}=\frac{-42-42\sqrt{2}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+42\sqrt{2}}{2*9}=\frac{-42+42\sqrt{2}}{18}
$